MISSING

PROPERTY OF ELLIPSE

A

MATHEMATICAL INVESTIGATION

BY

N. SAMBASIVA RAON. SAMBASIVA RAON. SAMBASIVA RAON. SAMBASIVA RAO email id : srnannuri@yahoo.co.in mobile : 9849592007

Article-0: It is not surprising, but interesting to see some similarities in the properties

of parabola and ellipse, for both are conics. 1. The locus of the feet of the perpendiculars drawn from the focus on the

tangents of the parabola is the tangent at the vertex. The corresponding locus in ellipse is the auxiliary circle.

2. The locus of the point of intersection of perpendicular tangents to a parabola is the directrix. The corresponding locus in ellipse is the director circle.

This suggests that as the ellipse gradually transforms into a parabola, its auxiliary circle transforms into the tangent at the vertex of the parabola and its director circle transforms into the directrix of the parabola. (As Lord Vishnu incarnates as Rama, Lakshmi takes birth as Sita and Adiseshu borns as Lakshmana).

The parabola may be regarded as an infinite ellipse whose second focus and hence the centre are at infinity on its axis. Therefore, both the auxiliary circle and director circle have become straight lines because their centres are at infinity and radii are infinity.

A light ray passing through one focus of an elliptical mirror, after reflection passes through the other focus. A light ray passing through the focus of a parabolic mirror, after reflection goes parallel to the axis of the parabola. This gives further strength to the above idea of a parabola as an infinite ellipse with second focus at infinity.

3. The circle drawn on the focal distance of any point on a parabola as diameter touches the tangent at vertex. In ellipse, we expect the circle to touch the auxiliary circle. And in fact, it touches the auxiliary circle. No surprise.

4. In a parabola, the circle drawn on any focal chord as diameter touches the directrix. In ellipse, we expect the circle to touch the director circle. But, Alas!, this property is conspicuously absent in an ellipse. Surprisingly, the circle neither touches the director circle nor the directrix. What happened to the property!!! It is a mystery.

In Article 1, we prove mathematically that the ellipse actually evolves into a parabola as 1e→ and its auxiliary circle and director circle evolve into the tangent at vertex and directrix respectively of the parabola. We prove that the directrix of ellipse also evolves to become the directrix of parabola. (Remember Darwin’s theory of Evolution!)

In Articles 2,3 and 4, we consider certain important properties of parabola and prove that they are inherited to the parabola by some corresponding properties of the ellipse.

In Article 5, we prove a property of the ellipse which is, most probably, hitherto unknown.

In Article 6, we unravel the mystery behind the Missing Property of Ellipse using the property of ellipse proved in Article 5. We prove that the Missing Property of Ellipse is not actually missing. It is very much present in Ellipse and we were not able to see it readily.

Article-1: Let E be the ellipse whose vertex is ( )0,0 , corresponding focus is ( ),0a and

with variable eccentricitye . Let P be the parabola 2 4y ax= . Find the equation of E and show that (i)

1eLt E P→

=

(ii) 1e

Lt→(Left directrix of E ) = Directrix of P

(iii) 1e

Lt→(Director circle of E ) = Directrix of P

(iv) 1e

Lt→(Auxiliary circle of E ) = Tangent at vertex of P

(v) 1e

Lt→ (Tangent at left vertex ofE ) = Tangent at vertex of P

Proof:

�

( ),P x y

( ),0S a,0ae−

0a

xe

+ =

•

•

•

•

•

•

ZO

( )0,0X

Y

M

Let Z be the foot of the left directrix of E . O is a point on the ellipse

,0SO a

e ZOZ e

− ⇒ = ⇒ =

∴ The equation of left directrix of E is 0a

xe

+ = .

For ellipse E , 2 2 2.SP e PM=

⇒ ( ) ( )2

2 22 2 ax a y e x ex a

e − + = + = +

2 2 2 2 2 22 2x ax y a e x aex a⇒ − + + = + +

⇒ ( ) ( )2 2 21 2 1 0e x a e x y− − + + = ( )1→

⇒ ( )( )

( )( )

222 2 2

2

121 0

1 11

a eax ae x y

e ee

+− − + + = +

− −−

⇒ ( )

( )( )( ) ( )

2 22 2

22 2

1

1 1 1 1 1

a ea y ax

e e e e e

+ − + = = − − − − −

⇒

2

2

2 2

1 11

1 1

ax

yea e

ae e

− − + = + − −

is the equation of ellipse E

(i) ( ) ( )2 2 2 2

11 2 1 0 4

eLt e x a e x y y ax→ − − + + = = =

1e

Lt E P→

⇒ = (using equation (1) for ellipse E )

(ii) [ ]1

0 0e

aLt x x a

e→

+ = = + =

⇒ 1e

Lt→(Left directrix of E ) = Directrix of P

(iii) 2 2

2 2

1

10

1 1 1e

a a eLt x y a

e e e→

+ − + − − = − − −

= 2 2 2

1

2 10

1 1e

a eLt x y x a

e e→

+ + − − = − −

= ( ) ( ) ( )2 2 2

11 2 1 0

eLt e x y ax a e→ − + − − + =

= ( ) [ ]0 2 0 0a x a x a− + = = + =

⇒ 1e

Lt→(Director circle of E ) = Directrix of P

(iv) 2 2

2

10

1 1e

a aLt x y

e e→

− + − = − −

= ( ) ( )2 2 2 2

1 1

20 1 2 0

1e e

axLt x y Lt e x y ax

e→ →

+ − = = − + − = −

[ ]0x= =

⇒ 1e

Lt→(Auxiliary circle of E ) = Tangent at vertex of P

(v) [ ] [ ]1

0 0eLt x x→

= = =

⇒ 1e

Lt→ (Tangent at left vertex ofE ) = Tangent at vertex of P

Article-2: Show that all chords of E that subtend a right angle at origin are concurrent at a

point Q and ( )1

4 , 0eLt Q a→

= [Here E refers to the ellipse of Article 1. Also note

that in the parabola P of Article 1, all chords subtending right angle at the vertex are concurrent at ( )4 , 0a ]

Proof:

O

A

B

•

•

•

X•Q

Let AB be 1lx my+ = Homogenizing the equation of ellipse E with the line AB , we get the equation to

the pair of lines ,OA OB .

( ) ( ) ( )2 2 21 2 1 0e x y a e x lx my− + − + + = using equation (1) of Article 1 for

ellipse E OA OB⊥

⇒ coefficient of 2x + coefficient of 2 0y =

⇒ ( ) ( )2 21 1 2 1 0 2 1 2e a e l l a e e− + − + = ⇒ + = −

⇒ ( )

2

2 11

2

a el

e

+ = −

⇒ ( ) ( )2

2 10 1

2

a el m

e

+ + = −

1lx my⇒ + = always passes through ( )

2

2 1, 0

2

a eQ

e

+ = −

( ) ( )

1

2 1 1,0 4 , 0

2 1e

aLt Q a→

+ ∴ = = −

Article-3: Let ( ),R h k be any point on E. Find subtangent and subnormal atR . Also show

that (i)

1eLt→(subtangent at R ) = 2h

(ii) 1e

Lt→ (subnormal at R ) = 2a [Here E refers to the ellipse of Article 1.

Also note that for parabola P of Article 1, the subtangent at any point is twice the abscissa of the point and subnormal is a constant equal to the semilatusrectum]

Proof:

T ( ), 0h

( ),R h k

GNormal

Tangent

•

•

•

Y

X

( ) ( )2 2 21 2 1 0e x y a e x− + − + = is the equation (1) of Article 1 for ellipse E

Differentiating w.r.t. x , we get

( ) ( )22 1 2 2 1 0dy

e x y a edx

− + − + =

Substituting ( ),h k , we get

( )( )

( )2

,

2 1 2 2 1 0at h k

dye h k a e

dx − + − + =

⇒ ( )

( ) ( ) ( ) ( )( )2

,

2 1 2 1 1 1

2at h k

a e e h e a e hdydx k k

+ − − + − − = =

∴ Tangent at ( ),h k is ( ) ( )( ) ( )1 1e a e h

y k x hk

+ − −− = −

Putting 0y = , we get

( ) ( )( )2

1 1k

x he a e h

−− =

+ − −

⇒ ( ) ( )( ) ( )( )

( )( ) ( )( )

2 2 21 1 1

1 1 1 1

ah e h e k ah ex

e a e h e a e h

+ − − − − += =

+ − − + − −

( )( ) ( )

( ) ( )

2 2 2

2 2 2

,

1 2 1 0

1 2 1

R h k lies on ellipse E

e h k a e h

h e k ah e

⇒ − + − + = ⇒ − − − = − +

⇒ ( )1

ahx

e h a=

− − where x is the abscissa of T in the figure

∴ 1e

ahLt x h

a→= = −−

1e

Lt→

⇒ (subtangent at R ) = 2h

Normal at ( ),h k is ( ) ( )( ) ( )1 1

ky k x h

e a e h−

− = −+ − −

Putting 0y = , we get

( ) ( )( )1 1e a e h x h+ − − = −

⇒ ( ) ( )( )1 1x h e a e h= + + − − where x is the abscissa of G in the figure

⇒ subnormal at ( ) ( )( )1 1R e a e h= + − −

∴ 1e

Lt→(subnormal at R ) = ( )2 0 2a a− = .

Article-4: Show that there exists a point K on the major axis of E , having the property that

for any chord PQ passing through 2 2

1 1,KPK QK

+ is a constant. Also show that

( )1

2 , 0eLt K a→

= [Here E refers to the ellipse of Article 1. Also note that the

parabola P of Article 1 has this property for the point ( )2 ,0K a ] Proof:

Q

θ( ),0K λ

P

Y

X

•

•

•

•O

Any point on the line PQ at a distance r from K will be in the form

( )cos , sinr rλ θ θ+ and if it lies on the ellipse E , then

( )( ) ( )( )22 2 21 cos sin 2 1 cos 0e r r a e rλ θ θ λ θ− + + − + + = (using equation (1) of

Article 1 for ellipse E )

( )( ) ( ) ( )( ) ( ) ( )2 2 2 2 2 2 21 cos sin 2 cos 1 2 1 cos 1 2 1r e e a e r e a eθ θ λ θ θ λ λ− + + − − + + − − +

= 0

( )( ) ( ) ( )( ) ( ) ( )( )2 2 2 21 cos sin 2 1 1 cos 1 1 2 0r e r e e a e e aθ θ λ θ λ λ− + + + − − + + − − =

,PK QK− are roots of this quadratic equation in r

⇒( ) ( )

( )

2

22 2

21 1 PK QK PK QKPK QK PK QK

− − −+ =

−

�

�

( ) ( )( )

2

22

2 2

2

22

b csumof roots product of roots a a

cproduct of rootsa

−−= =

( )2

22

20

b acreferring toax bx c

c−

= + + =

( ) ( )( ) ( ) ( )( ) ( )( )( ) ( )( )

22 2 2 2 2

222

4 1 1 cos 2 1 1 2 1 cos sin

1 1 2

e e a e e a e

e e a

λ θ λ λ θ θ

λ λ

+ − − − + − − − +=

+ − −

is a constant ⇒ coefficients of 2cos θ and 2sin θ must be equal and

Coefficient of ( ) ( )( ) ( ) ( )( )( )222 2cos 4 1 1 2 1 1 2 1e e a e e a eθ λ λ λ= + − − − + − − −

Coefficient of 2sin θ ( ) ( )( )2 1 1 2e e aλ λ= − + − −

( ) ( )( ) ( ) ( )( )( )22 24 1 1 2 1 1 2 1 1e e a e e a eλ λ λ⇒ + − − = + − − − −

⇒ ( ) ( ) ( )( )22 22 1 1 2 1e e a e aλ λ+ − − − + ( )( )( )2 21 2e a eλ λ= − − −

⇒ ( )( ) ( )( ) ( ) ( )( ) ( )22 2 2 22 1 1 1 2 2 1 1 2 1 0e e e e a e e e a eλ λ+ − + − − − + + + + =

⇒ ( )( ) ( ) ( )2 2 2 21 2 2 2 2 1 0e e a e a eλ λ− − − − + + =

⇒ ( )

( ) ( )2

22

2 120

1 1 2

a eae e e

λ λ+

− + =− − −

( )

( )( )( )( )

2 22

2 2

2

2

2 1

1 1 21

2 111 1 2

a ea ae e ee

eae e e

λ+

⇒ − = − − − − −

+ = − − − −

( ) ( )

2 2 2

2 21 1 2

a a ee e e

λ ⇒ − = − − −

( ) 21 1 2

a aee e e

λ⇒ = ±− − −

(Note: - K has two positions on the major axis of

E such that their midpoint is the centre of E )

2

11 2

a ee e

λ

⇒ = − − −

taking the left position of K

2

2

21 2

a e ee e

λ − −

⇒ = − −

( )( )

( )

( )2

2 2 2 2

2 1 2 11 2 2 2 2

e a eae e e e e e e

λ− +

⇒ = = − − − + − − +

( )

( )1

2 22

1 1 1e

aLt aλ→

∴ = =+

( )

12 , 0

eLt K a→

∴ =

Article-5:

RIGHT and LEFT ASSOCIATE CIRCLES of ELLIPSE: - A circle is drawn with right latus rectum as diameter. Another circle is drawn with centre on major axis such that it touches the above circle & auxiliary circle both internally. This circle is called right associate circle of ellipse. Similarly the left associate circle is also defined.

All the circles drawn on the right focal chords as diameters touch the right associate circle and similarly all the circles drawn on the left focal chords as diameters touch the left associate circle.

Proof:

A′

( )P α 2

,b

aea

S( ), 0ae

A M

( )Q β

• •

•

•

•

•

•X

Y

•O

L

Here let us consider the ellipse 2 2

2 2 1x ya b+ = . Let the circle drawn on right

latusrectum as diameter cut the major axis at M on the right side of the focus S .

( )( )2

2,0 1 , 0b

M ae a e ea

= + = + −

using ( )2 2 21b a e= − and ( ),0A a′ = −

Right Associate Circle

Centre 1C= =Midpoint of A M′ = ( )1

, 02

ae e−

Radius 1r= = 2

A M′= ( )222a

e e+ −

Circle with PQ as diameter

Equation of chord cos sin cos2 2 2

x yPQ is

a bα β α β α β+ + − + =

It passes through ( ) ( ), 0 cos cos 22 2

ae eα β α β+ −

⇒ = →

Centre 2 cos cos , sin cos2 2 2 2

C a bα β α β α β α β + − + − = =

( )( ) ( )1 cos , sin2 2ae beα β α β = + + +

using equation (2)

Radius 2

cos cos2 2

SP SQ a ae a aer

α β+ − + −= = = (using the formula for right

focal distance SP a ex= − )

( )cos cos cos cos2 2 2ae

a a aeα β α βα β + − = − + = −

( )( )2

2 2cos 1 cos2 2

aea ae a

α β α β+ = − = − + +

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 222 2 2

1 2

2 2 2 22 2 2

2 22 2 2 2 2

, cos sin4 4

cos 1 sin4 4

2 cos cos sin sin4

a e b eNow C C d e

a e a ee e

a ee e e

α β α β

α β α β

α β α β α β α β

= = + + + +

= + + + − +

= + + + + + + − +

( ) ( )2 2

2 2cos 2 cos 14

a ee eα β α β = + + + +

⇒ ( )1 cos2ae

d e α β= + +

( )

( )( )

2 2 2

1 2

1 2

cos2 2 2 2

1 cos2

ae ae ae aeAnd r r a a

aee

d r r

α β

α β

− = + − − + + +

= + +

⇒ = −

∴ All the circles drawn on the right focal chords as diameters touch the right associate circle and similarly all the circles drawn on the left focal chords as diameters touch the left associate circle.

Article-6:

Find the equation of the left associate circle of the ellipse E of Article 1 and show that

( )1e

Lt Left Associate Circle of E Directrix of P→

= where P is the parabola of

Article 1 Proof:

•

L

( ),0ae−M O

x a=

,01a

Ce

−

2, 0

1ae

−

• • • •

•

•L′

A

X

Y

( ),0S a

Ellipse E

( ) ( )2 2 21 2 1 0e x y a e x− + − + = from equation (1) of Article 1

Put x a= in the above equation to get L ( ) ( )2 2 2 21 2 1 0e a y a e⇒ − + − + =

( ) ( )( ) ( )( )( ) ( )

( )

2 2 2 2

2 2

22 2

2 1 1

1 2 1

1

, 1 , 1

1

y a e a e

y a e e

y a e

L a a e and L a a e

SL a e

⇒ = + − −

⇒ = + − +

⇒ = +

′⇒ = + = − +

⇒ = +

Let the circle on the left latusrectum LL′ as diameter cut the major axis at M on the left side of S .

SL SM⇒ =

( ), 0M ae⇒ = − and right vertex of 2

,01

aE A

e = = −

.

Left Associate Circle of E (Circle on MA as diameter)

2

2 2 2 20

1 1a a e

x y ae xe e

+ − − − = − −

⇒ ( ) ( ) ( )2 2 2 21 2 2 0e x y a e e x a e− + − − + − =

1eLt→

∴ (Left Associate Circle of E ) 22 2 0ax a = − − = [ ]0x a= + = = Directrix of P .