06 Polar Coordinates

8122019 06 Polar Coordinates

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CHAPTER 6

POLAR COORDINATES

TOPICS

1Relation Between Polar Coordinates And Cartesian Coordinates

2Distance Between Two Points

3Area Of The Triangle

4Equation Of Straight Line- Various Forms

5Equation Of The Circle

6Equation Of The Conic



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POLAR COORDINATES

POLAR COORDINATE SYSTEM

Let O be a fixed point in a plane and Ox

be a fixed ray in the plane With respect to these

two we can determine the position of any point in the plane The fixed point O is called pole and

the fixed ray Ox

is called initial line or polar axis Let P be a point in the plane such that OP = r IfangPox = θ then r θ are called polar coordinates of P The point P is denoted by (r θ) The non-

negative real number r is called radial distance the vector r OP=

is called radius vector and the

angle θ is called vectorial angle of the point P

RELATION BETWEEN POLAR AND CARTESIAN COORDINATES

Let P(x y) be a point in the Cartesian coordinate plane Take the origin O as pole and the

positive direction of x-axis as polar axis (initial line) Let (r θ) be the polar coordinates of P Then2 2

r OP x y= = +

x ycos sin

r rθ = θ = Thus x r cos y r sin= θ = θ

Note Conversion of Cartesian coordinates into polar coordinates is 2 2r x y = + x

cos r

θ =

ysin

rθ =

983120

θ 983119 983160

983154



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DISTANCE BETWEEN TWO POINTS

The distance between points (r1 1) (r2 2) is2 2

1 2 1 2 1 2r r 2r r cos( )+ minus θ minus θ

Proof

Let A(r1 θ1) B(r2 θ2) be the polar coordinates of two points wrt to O and OX

r1

r2

Then OA = r1 OB = r2 angAOx = θ1 angBOx = θ2

From ∆OAB2 2 2

2 21 2 1 2 1 2

AB OA OB 2OA OBcos AOB

r r 2r r cos( )

= + minus sdot ang

= + minus θ minus θ

there4 AB = 2 21 2 1 2 1 2r r 2r r cos( )+ minus θ minus θ

AREA OF THE TRIANGLE

The area of the triangle formed by the points

(r1 θ1) (r2 θ2) (r3 θ3) is 1 2 1 2

1r r sin( )

2Σ θ minus θ

983105

983106

θ983090

983119 983160

θ983089





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2 Taking origin as the pole and the positive x-axis as initial ray Convert the following

Cartesian equation into polar forms

Sol i) x2 + y

2 = a

2

The Relations between polar and cartesian coordinates are x=rcosθ y = rsinθ

r2

cos2

θ + r2

sin2

θ = a2

r2(cos2 θ + sin2 θ) = a2

ie r2 = a2

ii) y2 = 4ax


(r sinθ)2 = 4a(r cos θ)

r2 sin

2 θ = 4ar cos θ

2

cosr 4a

sin

θ=

θ

cos 1r 4a 4a cot csc

sin sin

θrArr = = θ θ

θ θ

iii)2 2

2 2

x y1

a b+ =


2 22 2

2 2

cos sinr r 1

a b

θ θ+ =

2 22

2 2cos sinr 1

a b θ θ+ =

iv) x2 ndash y

2 = a

2

r2 cos2 θ ndash r2 sin2 θ = a2

r2(cos2 θ ndash sin2 θ) = a2

r2 cos 2θ = a

2

3 Find the distance between the following pairs of points with polar coordinates

Sol i) Given points2

P 2 Q 4

6 3

π π

Distanct between the points is 2 21 2 1 2 2 1PQ r r 2r r cos( )= + minus θ minus θ

= ( )4 16 ndash 2 2 4 cos 120 ndash 30= + times times deg deg ( )20 ndash 16 cos 90 20 ndash 16 0= deg =

PQ 20 2 5= = units



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ii)7

P 3 Q 44 12

π π

Ans 13 units

iii) Given points P a Q 4a2 6

π π

PQ 13 a= units

II

1 Show that the points with polar coordinates

(0 0) (3 2) and (3 6) form an equilateral triangle

Sol Vertices of the triangle are A(0 0) B(3 π 2) and C(3 π 6)

2 2

1 2 1 2 2 1AB r r 2r r cos( )= + minus θ minus θ

0 9 ndash 2 0 3 cos (0 ndash p 2)= + times times 9 3= =

BC = ( )9 9 ndash 2 3 3 cos 90 ndash 30= + times times deg deg

=1

18 ndash 18 18 ndash 9 32

= times = =

CA = = 9 0 ndash 2 3 0 cos (p 6) 9 3+ times times = =

rArr AB = BC = CA

ABC is an equilateral triangle

2 Find the area of the triangle formed by the following points with polar coordinates

i)2

(a0) 2a 3a3 3

π π θ + θ +

ii) ( )2 5

5 4 003 6

π π minus minus

Sol i) vertices of the triangle are2

A(a0)B 2a C 3a3 3

π π θ + θ +

Area of the trangle is 1 2 1 2

1r r sin( )

2Σ θ minus θ

2 2 21 2 22a sin 6a sin 3a sin

2 3 3 3 3

π π π π = θ + minus θ + θ + minus θ minus + θ minus minus θ

1

2= [2a2sin(60deg) + 6a2sin(60deg) + 3a2sin(120deg)]



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2 2 21 3 3 32a 6a 3a

2 2 2 2

minus= sdot + minus + minus

2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus

are the vertices of the triangle

Ans 10 squnits

POLAR EQUATION OF A STRAIGHT LINE

The polar equation of a line passing through pole and making an angle α with the initial line

is = α

Proof

If P(r θ) is a point in the line then angPOX= θ and hence θ = α

Conversely if P(r θ) is a point such that θ = α then P lies in the line

there4 The equation to the locus of a point P in the line is θ = α

there4 The polar equation of the line is θ = α

THEOREM

The polar equation of a line passing through the points (r1 1) (r2 2) is

2 1 2 1

1 2

sin( ) sin( ) sin( )0

r r r

θ minus θ θ minus θ θ minus θ+ + =

Proof Let A(r1 θ1) B(r2 θ2) and P(r θ)

Then P lies in the line AB

hArr Area of ∆PAB = 0

1 2 1 1 2 1 2 2 2

1r r sin( ) (r r )sin( ) (r r)sin( ) 0

2θ minus θ + θ minus θ + θ minus θ =

hArr [ ]1 1 2 2 1 2 2 1r r sin( ) r sin( ) r r sin( )θ minus θ minus θ minus θ = θ minus θ

983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



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there4 The polar equation of line ie the locus of P is [ ]1 1 2 2 1 2 2 1r r sin( ) r sin( ) r r sin( )θ minus θ minus θ minus θ = θ minus θ

rArr 2 1 2 1

1 2


r r r


Note The polar equation of a line passing through the points (r1 θ1) (r2 θ2) is

[ ]1 1 2 2 1 2 2 1r r sin( ) r sin( ) r r sin( )θ minus θ minus θ minus θ = θ minus θ

COROLLARTY

The polar equation of a line passing through the pole and the point (r1 θ1) is θ = θ1

THEOREM

The polar equation of a line which is at a distance of p from the pole and whose normal

makes an angle α with the initial line is r cos ( ndash α) = p

Proof Let O be the pole of Ox

be the initial line

there4 ON = p angNOx = α

P(r θ) is a point on the line

hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p

rhArr θ minus α = hArr θ minus α =

Note If p = 0 then rcos (θ ndash α) = prArr rcos (θ ndash α) = 0rArr θ ndash α = π 2rArr θ = π 2 + α rArr θ = θ1

which is a line passing through pole

Note r cos (θ ndash α) = p

rArr r(cos θ cos α + sin θ sin α) = p

rArr (r cos θ) cos α + (r sin θ) sin α = p

rArr x cos α + y sin α = p (normal form in Cartesian system)

NOTE

The equation of a straight line which is at a distance of p units from pole and perpendicular to theinitial ray is r cos θ = p

NOTE

The equation of a straight line which is at a distance p units from the pole and parallel to the initial

ray is r sin θ = p

983119983160

α

983152



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THEOREM

The polar form of the line ax + by + c = 0 is a cos + b sin = kr where k = ndashc

THEOREM

The polar equation of a line parallel to the line a cos + b sin = kr is 1k a cos bsinr

θ + θ =

THEOREM

The polar equation of a line perpendicular to the linek

a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE

The polar equation of a line perpendicular to the line 1r cos( ) p is r sin( ) pθ minus α = θ minus α =

EXERCISE --- 6(B)

1 Find the polar equation of the line joining the points (5 2) and (ndash5 6)

Sol Given points are A(5 π 2) and B(ndash5 π 6)

Equation of the line joining A(r1 θ1) B(r2 θ2) is

2 1 2 1

1 2

sin( ) sin( ) sin( )

0r r r

θ minus θ θ minus θ θ minus θ

+ + =

sin sin sin6 2 6 2

0r 5 5

π π π π minus θ minus minus θ

+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus

θ π minus + minus = rArr θ minus minus θ =

5 3

sin cos6 2r

π

rArr θ minus minus θ =



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2 Find the polar coordinates of the points of intersection of the lines2

sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ

Sol Given lines are2

sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos

r= θ minus θ hellip(2)

Eliminating lsquorrsquo we get

2 sin sin 3sin 3 cos3

π θ + + θ = θ minus θ

2 sin cos cos sin sin 3sin 3 cos3 3

π π θ sdot + θ sdot + θ = θ minus θ

1 32sin 2cos 2sin 3sin 3 cos

2 2θ + θ + θ = θ minus θ 3sin 3 cos 3sin 3 cosθ + θ = θ minus θ

2 3 cos 0 cos 02

πθ = rArr θ = rArr θ =

Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3

π π= minus = sdot minus sdot = rArr =

Point of intersection is4

P 3 2

π

3 Find the polar equation of a straight line with intercepts lsquoarsquo and lsquobrsquo on the rays = 0 and

= 2 respectively

Sol Given line is passing through the points

A(a 0) and B(b π 2)

Equation of the line is

2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



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1 cos sin 1 cos sin0

r a b r a b

1 bcos a sin

r ab

θ θ θ θminus minus = rArr = +

θ + θ=

r(b cos θ + a sin θ) = abII

1 Find the polar equations of the lines passing through (2 4)

(i) parallel to (ii) perpendicular to the straight line7

4cos 3cosr

= θ + θ

Sol i) Given line is7

4cos 3sinr

θ + θ =

Equation of the line parallel to this line isk

4cos 3sinr

θ + θ =

This line is passing through A 24

π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =

Equation of the parallel line is

7 24cos 3sinr

θ + θ =

ii) Equation of the lint perpendicular to this line is

k k 4cos 3sin 4sin 3cos

2 2 r r

prime primeπ π + θ + + θ = rArr minus θ + θ =


π

k k 1 1 14sin 45 3cos 45 4 3

2 2 2 2 2rArr minus deg + deg = rArr = minus + = minus

2k 2

2rArr = minus = minus

Equation of the perpendicular line is2 2

4sin 3cos 4sin 3cosr r

minus θ + θ = minus rArr θ minus θ =



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2 Two straight roads intersect at angle 60deg A car is moving away from the crossing at the

rate of 25 kmph at 10 AM Another car is 52 km away from the crossing on the other

and is moving towards the crossing at the rate of 47 kmph at 10 AM Then find the

distance between the cars at 11 AM

Sol Let P and Q be the positions of the cars at 11 AM and 0 be the pole

Let O be the intersection of the roads

Let A be the position of the second car at 10AMOA = 52km

Let o be the position of the 1st car at 10 am

Velocity of the car is 25 kmph and that of the second is 47 kmph Let PQ be the positions of

two cars at 11 am respectively

OP = 25 OQ = 5 angPOQ = 60deg

2 2 21 2 1 2from OPQ PQ r r 2r r cos POQ

625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang

= + minus sdot sdot deg

= minus sdot sdot =

PQ 525 5 21= = km

POLAR EQUATION OF THE CIRCLE

The polar equation of the circle of radius lsquoarsquo and having centre at the pole is r = a

Proof

Let O be the pole and Ox

be the initial line If a point P(r θ) lies in the

circle then r = OP = radius = a Conversely if P(r θ) is a point such that

r = a then P lies in the circle

there4 The polar equation of the circle is r = a

983120983080983154983084θ983081

983119983160θ



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THEOREM

The polar equation of the circle of radius lsquoarsquo and the centre at

(c α) is r2 ndash 2cr cos( ndash α) = a

2 ndash c

2

Proof


be the initial line

Let P(r θ) be any point on the circle

Centre of the circle C = (c α) and radius of the circle = r

there4 OC = c CP = r angPOx = θ angCox = α

From ∆OPC we get

2 2 2CP OP OC 2OP OC cos POC= + minus sdot sdot ang

rArr 2 2 2a r c 2rccos( )= + minus θ minus α

2 2 2r 2rccos( ) a crArr minus θ minus α = minus

there4 The locus of P is2 2 2r 2rccos( ) a cminus θ minus α = minus

there4 The polar equation of the circle is 2 2 2r 2rccos( ) a cminus θ minus α = minus

NOTE

The polar equation of the circle of radius lsquoarsquo and passing through the pole is r = 2a cos(θ ndash α)

where α is the vectorial angle of the centre

NOTE

The polar equation of the circle of radius lsquoarsquo passing through the pole and its diameter as the

initial line is r = 2a cos θ

NOTE

The polar equation of the circle of radius lsquoarsquo and touching the initial line at the pole is r = 2a sin θ

α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



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POLAR EQUATION OF A CIRCLE FOR WHICH (R1 1) AND (R2 2) ARE EXTREMITIES OF DIAMETER

Given A(r1 θ1) B(r2 θ2) are the ends of the diameter LET P(r θ) be any point on the circle

AB is the diameterrArr angAPB = 90deg

AP2 + PB2 = AB2 hellip(1)

AP2 = r2 + r12 ndash 2rr1 cos(θ ndash θ1)

PB2 = r2 + r22 ndash 2rr2 cos(θ ndash θ1)

AB2 = r2 + r22 ndash 2r1r2 cos(θ1 ndash θ2)

Substituting in (1)

2 2 2 21 1 1 2r r 2rr cos( ) r r+ minus θ minus θ + + 2 22rr cos( )minus θ minus θ

2 21 2 1 2 1 2r r 2r r cos( )= + minus θ minus θ

Equation of the required circle is [ ]21 2r r cos( ) cos( )minus θ minus θ + θ minus θ 1 2 1 2r r cos( )+ θ minus θ = θ

POLAR EQUATION OF THE CONIC

The polar equation of a conic in the standard form is 1 ecosr= + θ

Proof Let S be the focus and Z be the projection of S on the directrix

Let S be the pole and SZ be the initial lineLet SL = be the semi latus rectum and e be the eccentricity of the conic

Let K be the projection of L on the directrix

L lies on the conic =LS

eLK

=

LS eLK eSZ SZ erArr = rArr = rArr =

Let P(r θ) be a point on the conic

983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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Let M be the projection of P on the directrix and N be the projection of P on SZ

P lies on the conicSP

e PS ePMPM

rArr = rArr =

r eNZ e(SZ SN) e(SZ SPcos )

e r cose

ercos

rArr = = minus = minus θ

= minus θ

= minus θ

r recos r(1 cos ) 1 ecosr

rArr + θ = rArr + θ = rArr = + θ

there4 The equation of the conic is 1 ecosr

= + θ

Note If we take ZS as the initial line then the polar equation of the conic becomes 1 ecosr= minus θ

Note The polar equation of a parabola is 21 cos 2cos

r r 2

θ= + θrArr =

Note The polar equation of a parabola having latus rectum 4a is 2 22a2cos a r cos

r 2 2

θ θ= rArr =

Note

The equation of the directrix of the conic 1 ecosr

= + θ

is ecosr

= θ

Proof Equation of the conic is 1 ecosr= + θ

Let Z be the foot of the perpendicular from S on the directrixrArr SZ = e

Let Q(rθ) be any point on the directrix of the conic

rArr SZ = SQ cos θ rArr r cos ecose r

= θrArr = θ

which is the equation of the directrix of the conic



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EXERCISE ndash 6(C)

1 Find the centre and radius of the circle r2 ndash 2r(3cos + 4 sin ) = 39

Sol

Equation of the circle is2

r 2r(3cos 4sin ) 39minus θ + θ =

2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39

5 5rArr minus θ + θ =

Let cos α = 35 sin α = 45

2r 10r(cos cos sin sin ) 39rArr minus θ α + θ α =

2 4r 10r cos( ) 39 where tan

3minus θ minus α = α =

Comparing with r2 ndash 2cr cos (θ ndash α) = a2 ndash c2

we get

c = 5 α = tanndash1 (43)

a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8

2 Find the polar equation of the circle whose end points of the diameter are

32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π

are the ends of the diameter of the circle


AB is a diameter of the circlerArr APB 90= deg

there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



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2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4

π = + minus θ minus


2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4

π = + minus sdot θ minus

π π = + minus sdot minus

= 4 ndash 4 cos 90deg = 4 ndash 0 = 4

From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4

π π + minus θ minus + + minus θ minus =

2 32r 2 2r cos cos 0

4 4

π π rArr minus θ minus + θ minus =

22r 2 2r cos cos 04 4

π π rArr minus θ minus + minusπ + + θ =

2

2

2

2r 2 2r cos cos 04 4

2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2

π π rArr minus θ minus minus θ + =

π rArr minus θ sdot =

rArr minus sdot θ sdot =

22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin

rArr minus θ = rArr minus θ =

minus θ = rArr = θ

Equation of the required circle is r = 2 sin θ

3 Find the polar equation of the circle with its centre at (4 6) and radius 5 units

Sol Centre is c 4 r 56π =

there4 c = 4 α = π 6 a = 5

Equation of the circle with centre (c α) and radius lsquoarsquo is

2 2 2r 2cr cos( ) c aminus θ minus α + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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2

2

r 8r cos 16 256

r 8r 9 06

π minus θ minus + =

π minus θ minus minus =

II

1 If PPprime and QQprime are the perpendicular focal chords of the conic prove that

1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol

Equation of two conic is 1 ecosr

= + θ

Given PPrsquo and QQrsquo are two perpendicular focal chords

Let P(SPθ

) where S is the focus then

1 1 ecos

1 ecosSP SP

+ θ= + θrArr =

Since PPrsquo is focal chord Pprime(SPprime 180deg + θ)

1 ecos(180 ) 1 ecosSP

= + deg + θ = minus θprime

1 1 ecos

SP

minus θ

=prime

PSPprime QSQprime are perpendicular focal chords

Coordinates of Q are SQ2

π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ

is a point on the conic

1 ecos 1 esinSQ 2

π = + + θ = minus θ

1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )

1 ecos 1 ecos 1 esin 1 esin

+ =prime prime

+ θ minus θ minus θ + θsdot + sdotsdot

2 2 2 2

2

1 e cos 1 e sinminus θ + minus θ=

=2

2

2 eminus

which is a constant

2 In a parabola prove that the length of a focal chord which is inclined at 6 to the axis is

four times the length of the latus rectum

Sol Equation of the conic is 1 ecosr = + θ

For a parabola e = 1

Equation of the parabola is 1 cosr

= + θ

and latusrectum is LLrsquo = 2

Let PPrsquo be the given focal chord

Let P SP6

π

then7

P SP 6

π prime prime

P and Prsquo are two points on the parabola therefore

3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP

= + deg = + deg + degprime

3 2 31 cos30 1

2 2

minus= minus deg = minus =

2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



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2 2PP SP SP

2 3 2 3prime prime= + = +

+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)

3 Prove that the perpendicular focal chords of a rectangular hyperbola are of equal length

Sol

Equation of the rectangular hyperbola is 1 2 cosr

= + θ

( ∵ For a rectangular hyperbola e 2= )

Let PPrsquo and QQrsquo be the perpendicular focal chords

Let P(SP α) then Pprime(SPprime 180deg + α) Q(SQ 90deg+α) and Qprime(SQprime 270deg+α)

Since P and Prsquo are points on the rectangular hyperbola

there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP

= + deg + α = minus αprime

SP1 2 cos

prime =minus α

PP SP SP1 2 cos 1 2 cos

prime = + = ++ α minus α

2

(1 2 cos 1 2 cos ) 2

cos21 2cos

minus α + + α minus= =

αminus α

there4 2

| PP |cos2

prime =α

hellip(1)

Q(SQ 90deg+α) is a point on the rectangular hyperbola there4

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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1 2 cos(90 ) 1 2 sinSQ

= + deg + α = minus α

SQ1 2 sin

=minus α

Qprime(SQprime 270deg+α) is a point on the rectangular hyperbola there4

1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α

QQ SQ SQ1 2 sin 1 2 sin

prime prime= + = +minus α + α

2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)

From (1) and (2) we get PPprime = QQprime

PROBLEMS FOR PRACTICE

1 Taking origin as the pole and the positive X-axis as the initial ray find the Cartesian

coordinates of point whose polar coordinates are (1 ndash 4)

Ans 1 1

2 2

minus

2 Taking origin as the pole and positive x-axis as initial ray find the polar coordinates of

the point p whose Cartesian coordinates are3 3

2 2

minus

Ans Polar coordinates of p are 34

π minus

3 Taking origin as pole and the positive x-axis as the initial ray convert the following polar

equation into Cartesian forms

i) r2 sin 2 = 18 Ans = xy = 9 ii) 5rsin cos

=θ minus θ

Ans = y x 5minus =

iii) r = 5 cos Ans x2 + y

2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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4 Taking the origin as the pole and positive

X-axis as initial ray convert the following Cartesian equations into polar equations

i) x2 ndash y

2 = 4y ii) y = x tan α

iii) x3= y

2(4 ndash x) iv) x

2 + y

2 ndash 4x ndash 4y + 7 = 0

Ans i) r cos 2θ = 4 sin θ ii) tanθ = tanα rArr θ = α iii)r cosθ = 4 sin2 θ

iv)r2 + 7 = 4r(cosθ + sinθ)


(0 0)7

5 518 18

π π

form an equilateral triangle

6 Find the area of the triangle formed by the points whose polar coordinates are

1 2 36 3 2

π π π

7 Find the polar equation of the straight line joining the points (3 3 4) and (2 4)

Ans 5r(sin cos ) 6 2θ minus θ =

8 Find the polar equation of the line passing through (3 3) and (i) parallel (ii) perpen-

dicular to the line3

3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =

9 Find the polar equation of a straight line passing through (4 20deg) and making an angle

140deg with the initial ray

Ans r cos( 50 ) 2 3sdot θ minus deg =

10 Find the polar coordinates of the point of intersection of the lines cos +2

cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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11 Find the foot of the perpendicular drawn from1

42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B

be foot the perpendicular drawn from A on the line1

sin cosr

θ minus θ =

there4 Equation of AB is

k sin cos

2 2 r

π π + θ minus + θ =

k

cos sin rθ + θ =

AB is passing through A1

42

π

k cos 45 sin 45

(1 2)rArr deg + deg =

1 1 1 1 1k 1

2 22 2 2

= + = + =

Equation of AB is 1cos sinr

θ + θ = hellip(1)

Equation of L is1

sin cosr

θ minus θ = hellip(2)

Solving (1) and (2)

cosθ + sinθ = sinθ ndash cosθ

2cosθ = 0rArr cosθ = 0rArr θ = π 2

1sin cos 1 0 1

r 2 2

π πthere4 = minus = minus =

r = 1

The foot of the perpendicular from A on L is B(1 π 2)

12Find the equation of the circle centered at (8120deg) which pass through the point

(4 60deg)

Ans r2 ndash 16r cos(θ ndash 120deg) + 16 = 0

983105

983106 983116

142

π



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13 Find the centre and radius of the circle2r 8r( 3 cos sin ) 15 0minus θ + θ + =

Ans Centre (c α) = 86

π

Radius a = 7

14 If PSQ is a chord passing through the focus S of a conic and l is the semi-latus rectum

show that1 1 2

SP SQ+ =

Sol

Let S be the focus and SX be the initial ray

Equation of the conic is 1 ecosr

= + θ

Let P(SP α) be a point on the conic

rArr 1 ecos 1 ecosr SP

= + αrArr = + α rArr 1 1 ecos

SP

+ θ=

hellip(1)

Q(SQ 180deg + α) is a point on the conic

rArr 1 ecos(180 ) 1 ecosSP


rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)

1 1 1 e cos 1 e cosSP SQ

+ α minus α+ = +

1 ecos 1 ecos 2+ α + minus α= =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

15 If PPprime QQprime are perpendicular focal chords of a conic show that1 1

aPP QQ

+ =prime prime

(constant)

Sol

Let Let PPrsquo and QQrsquo be the perpendicular focal chords of the conic 1 ecosr

= + θ


SP SP1 ecos 1 ecos

prime= =+ α minus α

SQ SQ1 esin 1 esin

prime= =minus α + α

PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α

QQ = SQ + SQprime 1 esin 1 esin

= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2

minus α minus α+ = +

prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

POLAR COORDINATES

POLAR COORDINATE SYSTEM

Let O be a fixed point in a plane and Ox

be a fixed ray in the plane With respect to these

two we can determine the position of any point in the plane The fixed point O is called pole and

the fixed ray Ox

is called initial line or polar axis Let P be a point in the plane such that OP = r IfangPox = θ then r θ are called polar coordinates of P The point P is denoted by (r θ) The non-

negative real number r is called radial distance the vector r OP=

is called radius vector and the

angle θ is called vectorial angle of the point P

RELATION BETWEEN POLAR AND CARTESIAN COORDINATES

Let P(x y) be a point in the Cartesian coordinate plane Take the origin O as pole and the

positive direction of x-axis as polar axis (initial line) Let (r θ) be the polar coordinates of P Then2 2

r OP x y= = +

x ycos sin

r rθ = θ = Thus x r cos y r sin= θ = θ

Note Conversion of Cartesian coordinates into polar coordinates is 2 2r x y = + x

cos r

θ =

ysin

rθ =

983120

θ 983119 983160

983154



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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Proof


r1

r2


From ∆OAB2 2 2

2 21 2 1 2 1 2


r r 2r r cos( )

= + minus sdot ang





(r1 θ1) (r2 θ2) (r3 θ3) is 1 2 1 2

1r r sin( )

2Σ θ minus θ

983105

983106

θ983090

983119 983160

θ983089





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



Sol i) x2 + y

2 = a

2


r2

cos2

θ + r2

sin2

θ = a2


ie r2 = a2

ii) y2 = 4ax



r2 sin

2 θ = 4ar cos θ

2

cosr 4a

sin

θ=

θ


sin sin

θrArr = = θ θ

θ θ

iii)2 2

2 2

x y1

a b+ =


2 22 2

2 2

cos sinr r 1

a b

θ θ+ =

2 22

2 2cos sinr 1

a b θ θ+ =

iv) x2 ndash y

2 = a

2



r2 cos 2θ = a

2



P 2 Q 4

6 3

π π



PQ 20 2 5= = units



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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ii)7

P 3 Q 44 12

π π

Ans 13 units


π π

PQ 13 a= units

II




2 2




=1


= times = =


rArr AB = BC = CA



i)2

(a0) 2a 3a3 3

π π θ + θ +

ii) ( )2 5

5 4 003 6

π π minus minus


A(a0)B 2a C 3a3 3

π π θ + θ +


1r r sin( )

2Σ θ minus θ


2 3 3 3 3


1




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2 21 3 3 32a 6a 3a

2 2 2 2


2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus


Ans 10 squnits



is = α

Proof





THEOREM


2 1 2 1

1 2


r r r





1 2 1 1 2 1 2 2 2




983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149




Proof


r1

r2


From ∆OAB2 2 2

2 21 2 1 2 1 2


r r 2r r cos( )

= + minus sdot ang





(r1 θ1) (r2 θ2) (r3 θ3) is 1 2 1 2

1r r sin( )

2Σ θ minus θ

983105

983106

θ983090

983119 983160

θ983089





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



Sol i) x2 + y

2 = a

2


r2

cos2

θ + r2

sin2

θ = a2


ie r2 = a2

ii) y2 = 4ax



r2 sin

2 θ = 4ar cos θ

2

cosr 4a

sin

θ=

θ


sin sin

θrArr = = θ θ

θ θ

iii)2 2

2 2

x y1

a b+ =


2 22 2

2 2

cos sinr r 1

a b

θ θ+ =

2 22

2 2cos sinr 1

a b θ θ+ =

iv) x2 ndash y

2 = a

2



r2 cos 2θ = a

2



P 2 Q 4

6 3

π π



PQ 20 2 5= = units



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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ii)7

P 3 Q 44 12

π π

Ans 13 units


π π

PQ 13 a= units

II




2 2




=1


= times = =


rArr AB = BC = CA



i)2

(a0) 2a 3a3 3

π π θ + θ +

ii) ( )2 5

5 4 003 6

π π minus minus


A(a0)B 2a C 3a3 3

π π θ + θ +


1r r sin( )

2Σ θ minus θ


2 3 3 3 3


1




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2 21 3 3 32a 6a 3a

2 2 2 2


2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus


Ans 10 squnits



is = α

Proof





THEOREM


2 1 2 1

1 2


r r r





1 2 1 1 2 1 2 2 2




983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



Sol i) x2 + y

2 = a

2


r2

cos2

θ + r2

sin2

θ = a2


ie r2 = a2

ii) y2 = 4ax



r2 sin

2 θ = 4ar cos θ

2

cosr 4a

sin

θ=

θ


sin sin

θrArr = = θ θ

θ θ

iii)2 2

2 2

x y1

a b+ =


2 22 2

2 2

cos sinr r 1

a b

θ θ+ =

2 22

2 2cos sinr 1

a b θ θ+ =

iv) x2 ndash y

2 = a

2



r2 cos 2θ = a

2



P 2 Q 4

6 3

π π



PQ 20 2 5= = units



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

ii)7

P 3 Q 44 12

π π

Ans 13 units


π π

PQ 13 a= units

II




2 2




=1


= times = =


rArr AB = BC = CA



i)2

(a0) 2a 3a3 3

π π θ + θ +

ii) ( )2 5

5 4 003 6

π π minus minus


A(a0)B 2a C 3a3 3

π π θ + θ +


1r r sin( )

2Σ θ minus θ


2 3 3 3 3


1




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2 21 3 3 32a 6a 3a

2 2 2 2


2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus


Ans 10 squnits



is = α

Proof





THEOREM


2 1 2 1

1 2


r r r





1 2 1 1 2 1 2 2 2




983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



Sol i) x2 + y

2 = a

2


r2

cos2

θ + r2

sin2

θ = a2


ie r2 = a2

ii) y2 = 4ax



r2 sin

2 θ = 4ar cos θ

2

cosr 4a

sin

θ=

θ


sin sin

θrArr = = θ θ

θ θ

iii)2 2

2 2

x y1

a b+ =


2 22 2

2 2

cos sinr r 1

a b

θ θ+ =

2 22

2 2cos sinr 1

a b θ θ+ =

iv) x2 ndash y

2 = a

2



r2 cos 2θ = a

2



P 2 Q 4

6 3

π π



PQ 20 2 5= = units



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

ii)7

P 3 Q 44 12

π π

Ans 13 units


π π

PQ 13 a= units

II




2 2




=1


= times = =


rArr AB = BC = CA



i)2

(a0) 2a 3a3 3

π π θ + θ +

ii) ( )2 5

5 4 003 6

π π minus minus


A(a0)B 2a C 3a3 3

π π θ + θ +


1r r sin( )

2Σ θ minus θ


2 3 3 3 3


1




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2 21 3 3 32a 6a 3a

2 2 2 2


2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus


Ans 10 squnits



is = α

Proof





THEOREM


2 1 2 1

1 2


r r r





1 2 1 1 2 1 2 2 2




983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

ii)7

P 3 Q 44 12

π π

Ans 13 units


π π

PQ 13 a= units

II




2 2




=1


= times = =


rArr AB = BC = CA



i)2

(a0) 2a 3a3 3

π π θ + θ +

ii) ( )2 5

5 4 003 6

π π minus minus


A(a0)B 2a C 3a3 3

π π θ + θ +


1r r sin( )

2Σ θ minus θ


2 3 3 3 3


1




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2 21 3 3 32a 6a 3a

2 2 2 2


2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus


Ans 10 squnits



is = α

Proof





THEOREM


2 1 2 1

1 2


r r r





1 2 1 1 2 1 2 2 2




983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

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42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2 21 3 3 32a 6a 3a

2 2 2 2


2 2 232a 6a 3a

4= + minus

2 23 5 3(5a ) a squnits

4 4= =

ii) ( )2 5

A 5 B 4 C 003 6

π π minus minus


Ans 10 squnits



is = α

Proof





THEOREM


2 1 2 1

1 2


r r r





1 2 1 1 2 1 2 2 2




983119983160

α

983119

983105

983106

983120

983160

983154983089

θ983090

θ983089

983154983090



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


rArr 2 1 2 1

1 2


r r r




COROLLARTY


THEOREM




be the initial line



hArr ON

cos NOPOP

ang =

pcos( ) r cos( ) p








NOTE


NOTE


ray is r sin θ = p

983119983160

α

983152



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM


THEOREM


θ + θ =

THEOREM


a cos bsinr

θ + θ = is

1k a cos bsin

2 2 r

π π + θ + + θ =

NOTE


EXERCISE --- 6(B)




2 1 2 1

1 2


0r r r


+ + =

sin sin sin6 2 6 2

0r 5 5


+ + =

minus

sin3 cos 1 36

0 sin cos2r 5 5 5 6 2r

π θ minus


5 3

sin cos6 2r

π




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


sin sinr 3

π = θ + + θ

and4

3sin 3 cosr

= θ minus θ


sin sinr 3

π = θ + + θ

hellip(1)

43sin 3 cos









2 3 cos 0 cos 02


Substituting in (1)

4 43sin 4cos 3 1 4 0 3 r

r 2 2 3



P 3 2

π


= 2 respectively




2 1 2 1

1 2


r r r


sinsinsin(0 )22 0

r a b

π πθ minus

minus θ + + =



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


r a b r a b

1 bcos a sin

r ab


θ + θ=




4cos 3cosr

= θ + θ


4cos 3sinr

θ + θ =


4cos 3sinr

θ + θ =


π

rArrk

4 cos 45 3sin 452

deg + deg =

rArr1 1 7

k 2 4 3 2 7 22 2 2

= + = sdot =


7 24cos 3sinr

θ + θ =



2 2 r r



π

k k 1 1 14sin 45 3cos 45 4 3


2k 2







983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149













625 25 2 25 5cos 60

1650 2 125 525

2

∆ = + minus ang


= minus sdot sdot =

PQ 525 5 21= = km



Proof






983120983080983154983084θ983081

983119983160θ



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

THEOREM



2 ndash c

2

Proof


be the initial line




From ∆OPC we get






NOTE



NOTE



NOTE


α

983107983080983139983084α983081

983119 983160

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149








Substituting in (1)










eLK

=



983123 983105983118 983130

983117

983115983116

983120

983154983148

θ

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



e PS ePMPM

rArr = rArr =


e r cose

ercos


= minus θ

= minus θ




= + θ



r r 2

θ= + θrArr =


r 2 2

θ θ= rArr =

Note


= + θ

is ecosr

= θ





= θrArr = θ




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

EXERCISE ndash 6(C)


Sol



2 23 4 5+ =

2 3 4r 2r5( cos sin ) 39







we get


a2 ndash c2 = 39rArr a2 = 39 + c2 = 25+39 = 64

a2 = 64rArr a = 8

centre is1 4

c 5 tan3

minus

radius a = 8


32 and 2

4 4

π π

Sol

Given3

A 2 B 24 4

π π




there4 AP2 + PB2 = AB2 -----(1)

983105 983106

983120983080983154983084θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2

2

AP r 2 2r 2 cos4

r 2 2 2r cos

4



2 2

2

3BP r 2 2r 2 cos

4

3AB 2 2 2 2 2 cos

4 4




From (1)

2 2 3

r 2 2 2r cos r 2 2 2r cos 44 4



4 4


22r 2 2r cos cos 04 4


2

2

2


2r 2 2r 2sin sin 04

12r 2 2r 2 sin 0

2




22r 4r sin 0 2r(r 2sin ) 0

r 2sin 0 r 2sin






there4 c = 4 α = π 6 a = 5





983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2

2

r 8r cos 16 256

r 8r 9 06



II


1 1 constant

(SP)(SP ) (SQ)(SQ )+ =

prime prime

Sol


= + θ


Let P(SPθ


1 1 ecos

1 ecosSP SP

+ θ= + θrArr =




1 1 ecos

SP

minus θ

=prime



π + θ

and

3Q are SQ

2

π prime prime + θ

Q SQ2

π + θ


1 ecos 1 esinSQ 2


1 1 e sin

SQ

minus θ=

3Q SQ

2

π prime prime + θ


31 ecos 1 esin

SQ 2

π = + + θ = + θ

prime

983120983121

983123

π983087983090

983120prime

θ

983121 prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 1 esin

SQ

+ θ=

prime

1 1

(SP)(SP ) (SQ)(SQ )


+ =prime prime


2 2 2 2

2


=2

2

2 eminus

which is a constant






= + θ



Let P SP6

π

then7

P SP 6

π prime prime


3 2 31 cos30 1

SP 2 2

+= + deg = + =

2SP

2 3rArr =

+

And 1 cos 210 1 cos(180 30 )SP


3 2 31 cos30 1

2 2


2SP

2 3prime =

minus

L

Lrsquo

983120

983123

983120prime



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

2 2PP SP SP


+ minus

2 (2 3 2 3)8 4(2 )

4 3

minus + += = =

minus

= 4 (latus rectum)


Sol


= + θ





there4 1 1 2 cos

1 2 cos

SP SP

+ α= + αrArr =

SP1 2 cos

=+ α

1 2 cos(180 ) 1 2 cosSP


SP1 2 cos

prime =minus α



2

(1 2 cos 1 2 cos ) 2

cos21 2cos


αminus α

there4 2

| PP |cos2

prime =α

hellip(1)


983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083

αθ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

1 2 cos(90 ) 1 2 sinSQ


SQ1 2 sin

=minus α


1 2 cos(270 )SQ

1 2 sin SQ1 2 sin

= + deg + αprime

prime= + α = =+ α



2

(1 2 sin 1 2 sin ) 2

cos21 2sin

+ α + minus α= =

αminus α

hellip(2)





Ans 1 1

2 2

minus



2 2

minus


π minus




=θ minus θ

Ans = y x 5minus =


2= 5x iv) 2

r cos a2

θ= Ans y

2 + 4ax = 4a

2



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



i) x2 ndash y


iii) x3= y

2(4 ndash x) iv) x

2 + y





(0 0)7

5 518 18

π π



1 2 36 3 2

π π π





3cos 4sinr

θ + θ =

Ans (i)3(3 4 3)

3cos 4sin2r

+θ + θ =

ii)3(4 3 3)

4cos 3sin2r

minusθ minus θ =





cos3 r

π θ minus =

and2

cos cos

3 r

π θ + θ + =

Ans P(43 0deg)



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


42

π

on the line1

sin cosr

θ minus θ =

Sol

Let B


sin cosr

θ minus θ =


k sin cos

2 2 r


k

cos sin rθ + θ =


42

π

k cos 45 sin 45


1 1 1 1 1k 1

2 22 2 2

= + = + =


θ + θ = hellip(1)

Equation of L is1

sin cosr


Solving (1) and (2)



1sin cos 1 0 1

r 2 2


r = 1



(4 60deg)


983105

983106 983116

142

π



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149

983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149



π

Radius a = 7


show that1 1 2

SP SQ+ =

Sol



= + θ




SP

+ θ=

hellip(1)




rArr1 1 e cos

SQ

minus α=

hellip(2)

Adding (1) and (2)


+ α minus α+ = +




983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081



983159983159983159983086983155983137983147983155983144983145983141983140983157983139983137983156983145983151983150983086983139983151983149


aPP QQ

+ =prime prime

(constant)

Sol


= + θ


SP SP1 ecos 1 ecos


SQ SQ1 esin 1 esin


PP = SP + SPprime

1 ecos 1 ecos

= ++ α minus α

2 2 2 2

(1 e cos 1 ecos ) 2

1 e cos 1 e cos

minus α + + α= =

minus α minus α


= +minus α + α

2 2 2 2

(1 esin 1 esin ) 2

1 e sin 1 e sin

+ α + minus α= =

minus α minus α

2 2 2 21 1 1 e cos 1 e sin

PP QQ 2 2


prime prime

22 e(constant)

2

minus=

983120983080983123983120983084α 983081

983080983123983120prime983084983089983096983088983083α 983081

983080983123983121 prime983084983090983095983088deg983083α 983081

983080983123983121 prime983084983097983088deg983083α θ983081

Date post:	03-Jun-2018
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06 Polar Coordinates

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