Ganguli, Theory of Plane Curves, Chapter 6

8/14/2019 Ganguli, Theory of Plane Curves, Chapter 6

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CHAPTER VI.POLAR RECIPROCAL AND OTHER DERIVED CUlWES.

113. In Chapter I, we have enunciated the generalprinciples of reciprocation.* The theory of point reciproca-tion has beer. fully discussed by Salmon in his Treatise onConic Sections, Chapter XV, and therefore it is not necessaryto enter into a detailed discussion of the theory in thepresent work. We shall here study the properties of polarreciprocal curves mainly with reference to their singularpoints. At the outset, however, it is useful to recall thefollowing definition:

Suppose we have a certain conic, called the base-conic.The envelope of the polar line of any point on the givencurve S is called the polar reciprocal S' of S with regard tothe conic. The relation between Sand S' is reciprocal, one~beingderived from the other by the same process.

Let C=0 be the base-conic, and 1=0 any given curve.Let the equation of a tangent to 1be written in the form

p=x cos a+y sin a (1)and the condition that this line touches the given curve.be written in the form

(2)If now (J/, y') be

the tangent (1) must(x', y'), i.e. with

the pole of this tangent w.r t. C=0,be identical with the polar line of

xf'. +si' y +zf', =0 (3). cosc _ -f'. sin a =I',p - 1', and p = f' ,

. Poncelet-Memoire sur la theorie generale des polaires" " , p " , q " , etc. O"ll~Bd. IV,("21)) p p . 1-71.


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1 3 6 THIWRY 01' PLANE CORVES

whence 1-1".+I" y -I'-- , and tan a_j'~'pi f', _Therefore, the locus of Ct', y'), i.e. the polar reciprocal

of j is r .+j" y = 1 " , : { < p ( tan - ';: ) 1 IThe base C=O may be any conic-a ci dIe for example,

and a point-circle, in particular; and in that case, thereciprocal is said to be taken w. r. t, the circle,

In this case we may define the polar reciprocal curve ina different manner as follows:

If OP be the perpendicular from the pole upon thetangent to a given curve, and if on OP or OP produceda point Q be taken such that OP. OQ =kt (const), the locusof Q is-called the reciprocal. polar of the given curve withregard to a circle of radius k and centre at O .

It is evident from this definition that the polar reciprocalcurve is the inverse of the first positive pedal of the curve,

Ex, 1. Find the polar reciprocal of the curve(x !a )m +(y /b)"'= l

with regard to 9 circle of radius k and centre at the origin,The condition that the line ,t COS a + I J sin a=p is a tangent to the

curve is-(a cos a ) m-l + (b sin a);;;::; =p

If now r denotes the radius vector of the reciprocal polar, we havep1'= k ', and substituting the value of p in the above condition, we obtain

m ,nm -l m - = J .(a cos a) + (b sin a) =(k'Mwhence _ ._ .. '" 2 '"(ax) "-1 + (by) m : : - , " " k ,;;::-;


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RECIPROCAL POLAR CUR.VES 137

If, in particular, 111 =2, the polar reciprocal of the ellipsex'ja'+y'jb'=l

. is found to be a';v' + b 'yO =k",Ea:. 2. Show that the polar reciprocal of the curve a:M y' =am+. with

regard to a circle whose centre is at the origin is another curve of thesame kind.E, 3. Show that the polar reciprocal of the curve j '' ' ' =a ' " cos m6

1M.t. the hyperbola ,.scos28=a is"':'+1 ( m ) 0 , . ~ IC08 -- =rt",+1

Ex. 4. Prove that the reciprocal of the same curve 117 . ,'. f. a circlewith centre at the pole is of the form in Ex. 3.

114. POLAR RECIPReCAL I : : : - r HmIOGE~EOUS CO-ORDINATES:Theorem: If c p a , "7, ~) =0 be the tangential equation of acurve, the point equa tt'on of its reciprocal polar with respect

to the imaginary circle : V I+ y 2 + z =0 is< / > ( : 1 ' , y, z)=0.

Let ~,I:+"7Y+~z=O be a tangent to the curve, andlet ( : 1 : ' , y', s') be its pole with respect to X2 +yt +zt =0.Therefore, the polar of ( : 1 : ' , s', z'), i.e., I r a , ' +yy' +z~'=O mustbe identical with ~x+"7y+~z=O.i.e., we must have-

and consequently, < / > ( : v ' , y', z') =0, i.e., the locus of the point(e', y' z') is the curve cp(.r, y, z)=O , which is the reciprocalpolar of the given curve.

Thus, if ?n be the degree of the tangential equation of acurve, the degree of its reciprocal polar curve is also m. Butthe degree of the tangen tial equation of a curve is equal to

:;. its class. *IISalmon, Conics, 321.


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l38 THEOll,Y OF PLANE CURVESTherefore, the degree of the reciprocal pola r of a CUlTC i

equal to its class, and vice Vel'SitThe conic x' +y' +Z2 =0 is called the base-conic.

It is to be noticed, in particular, that if the conditionthat the line ~.l l+y+~=O touches a curve be put into theform c f > a , n =0, i.e., if the tangential equation of the curvebe c f > a , ~ ) =0, then its polar reciprocal with regard tothe parabola ,1)' + 2y=0 is c f > ( . l ' , y) =0.

Em . 1. Find the reciprocal polar of the curve z( z 2 - xy) =xy(w + y)lV.I'. t m ' + y' + z' =0.

Em , 2. Prove that the polar reciprocal of cp(~, '7, 0=0 with respectto the base-conic

aw ' + by' 4 - cz + 2fyz + 2gzw + 2lw;y =0is cp(a ,v+hy+gz, hx+bY+ fz, gx+ fy+cz)= .O .Ex. 3. Show that a conic and its reciprocal have, in general, a

common self-conjugate triangle with the base-conic.

115. TANGENTIAL EQl)ATION DERIVED FROM POINT EQl)ATION :Let f(,f, y, z )==O be the equation of a curve, and let

~.ll+1JY+~z=O be a tangent to it at the point ( . 1 " , y', z').Then this line must be identical with

which is the equation of the tangent at ( ; 1 " , y ', c').:. We have ~: Y J : ~=I' : l' 2 : l' 3 (1)Since (x', y ', t') is a point on f(a', y , z)=O, we have

f (x', y', z')=O (2)Eliminating (x', y', z ') between equations (1) and (2), weobtain the required tangential equation of the curve.


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,

RECIPROCAL POLAR CURVES 139But, by Euler's Theorem,

x' af +y' at +z' aj =nf(c' y' z')-O (3)a :Il' a y' a z' ',,-Therefore, the process is simplified, if elimination is

effected between (1) and (3). We thus obtain the tangentialeqnation in the form c p a , ' Y / , n =0.Second Method ;-

Eliminating one of the variables z (say) between theequation of the curvefCc, y, z)=O (1)

(2)ndwe obtain the equation of the lines -

(3)which join the vertex 0(0, 0, I) of the triangle of referencewith the intersections of (1) with (2).

If the line (2) touches the curve (1), two of the lines (3)will be coincident, and the equation (3), regarded as anequation in x/y, will have two equal roots. Hence thediscriminant of (8) will vanish, which gives the tangentialequation of the curve.

Ex. 1. Find the tangential equation of the cubic ,c' + y' + z" =0.First Method :-

We have-~'I:(=x' I:y" : z'!

and, also~'"'+ 'Iy' + (z'=.

But, x'=v~, y'= V'l, z'= ,1(.


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140 THEORY OJ!' PLANE CURVES

or, rationalising, we obtai n-


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~ECIPROCAL POLAR CURVES 141

Consequently, the discriminant f(,v, y, z) must vanish,which gives f(,r, y, z) =0 as the point equation of thecurve.Second M ethod :-

Let (.u , y, z ) be a point, and t, Y J ' , " be the co-ordinates ofits polar line. But the pole of the line (t', Y J ' , nis-

'C : y : z = ~ t :g : :~!,Also,

where m is the degree of c p , since the line a ', Y J ' , " )touches the curve.

Eliminating (t', r/, nbetween these equations, we obtain-the locus of (re, y, z) in the formf(x , y, z)=O .Ex. 1. Form the point equation of the curve whose tangential

equation is ~. + 1/" + (' =0.Eliminating' between this equation and ~x+ 1 /Y + (z= 0, we obtain

The discriminant of this is-

Z6(X 3 +y3_zS)=4ro3y'z " .But, evidently ~J)+1/Y+(z=O is not a point on the curve, if z=O ... The factor Z6 is irrelevant, a.ndthe point equation of the curve is

Thus the order of this curve, which is of class 3, is 6.

i f < C. A. Scott,-loc. cit., 64,


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142 THEORY OF PLA.NE CURVEs

E, 2. The point equation of 4~' + 27'1'(=0 is found to bex3 -y'z =0.

For, eliminating (between the given eqnation and ~J)+ '1!I+ (z=O ,we obtain-

the discriminant of which ;8 ",' - y',. Hence the point equation is",3='z, and tho order of this curve, which is of class 3, is also 3.

E .. 3. Find the point equation of curves defined by the followingtangential equations:-

(iii) 27(t''12(+4(a~+O'=0.117. By a comparison of the examples in 115, 116, it

is seen that the class of a cubic may be as much as 6, or aslittle as 3 ; and conversely, the degree of a curve of class 3may be either 6 or 3. Hence, in general, it follows that theclass of a carve does not depend solely on its order, nor' itsorder upon the class. Further, it is seen that if f( .c, y, z) =0is the point equation of a curve whose tangential equation is4 > a , ' 1 / , ' ) =0, then f(~ , ' 1 / , ') =0 is the tangential equation ofa curve whose point equation is 4 > ( . t " y , e)=0, and the classof a curve is the same as the degree of its tangentialequation.

Now, corresponding to the locus f( c , y, .)=0, there is,by the Principle of Duality,* an envelope f\~, '1 / , n = O , andthe point equation of this, found by the above method, is4 > ( . c , y , z) =0. Instead of discussing the two equationsf (.c, y, z )=O and 4 > a , '1 / , ,)=0 of the same curve, we mayconveniently consider two distinct curves f( .v, y, z )=O and4 > ( . e , y , z) =0. These two curves are called Polar reciprocalcurves. The relation between them is a reciprocal one, ie.,the une can be obtained from the other in the same way asthe latter is from the former : and the line and pointproperties of one are exactly the same as the point and lineproperties of the other .

Plucker=-" System der Aualytischen Geometrio (1835).


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RECIPROCAl, POI,AR CURVES 143118. THEOREM: A node on a curve corresponds to a double

tangent on the reciprocal. curve, and uice vers(/.Through a node there pass two branches of the curve,

with a distinct tangent to each. Thus the point may beregarded once as belonging to one branch and once to theother. Therefore, the polar lines of these two points aretwo corresponding tangents to the reciprocal curve, whichbecome ultimately coincident. The nodal tangents corres-pond to two distinct points on the reciprocal, the tangentsat which become ultimately coincident. Thus, the nodaltangents with their coincident points of contact reciprocateinto two distinct points on the reciprocal curve, with thetangents at those points coincident. Hence, a node corres-ponds to a tangent of the reciprocal curve, which touche s itat two distinct points, i,e., is a double tangent.

Note. Since a conjugate point is a real point with imaginarytangents, the polar line of the conjugate point is a real tangent withimaginary points of contact.

Hence a conjuqaie point on Q, curve cOl'l'esponds to a real. double tangentof the reciprocal C!tl've uiiih. imagilwl'Y point, of contact.

119. THEOREM: A C?lSP on a curve corresponds to aninfle rional tangent on thereciprocal curve and vice versa.

At a cusp, the two branches of the curve have a commontangent. Let P be the cusp, and PQ the cuspidal tangent.From any point Q on the tangent, we can draw two othertangents QT, QR, one to each branch of the curve. Thereciprocals of these three concurrent tangents will be threepoints on the reciprocal curve, which are collinear on thepolar of Q. Ultimately when Q moves up to coincidencewith P, the three tangents coincide with the cuspidaltangent, and the three corresponding points becomecollinear, i.e., three consecutive points on the reciprocalcurve are collinear. Therefore the tangent has a contactof the second order, and consequently, the cusp corresponds


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1 4 4 THEORY OF PLANE CURVES

Second proof:At a cusp the moving point turns back along the tangent,

i.e., it chang-es its sense of motion in the direction of thetangent. Therefore, in the reciprocal curve, the envelopingline changes its direction of motion about its point of contact,i.e., the line becomes stationary for a moment before thesense of its motion is changed. Therefore it is an inflexionaltangent, and the point of contact corresponds to the cuspidaltangent.

Thus we see that a cusp and an inflexional tangent arestationary elements, while the node and the double tangentare simply double elements. *

120. THEORE~I: If a WTt'e of order n hus 0 nodes, thedeqree of its reciprocal pola is n(n-l) - 28.

We have seen that the reciprocal of a curve of ordern is, in general, of degree n(n-l), this being the numberof tangents that can be drawn from any point to the curve,and this is equal to the number of intersections of thecurve with its first polar. If, however, the curve has anode, the first polar passes through it, and the point countsas two among the intersections of the curve with its firstpolar. But the line joining the node to the pole is not to bereckoned as a tangent. Thus the number of tangents whichcan be drawn from any point to the curve is diminished bytwo for every node on the curve. Therefore, if there are 8nodes, the number of tangents is diminished by 28, and,consequently this number reduces to n(n-l)-2o; or, 111other words-

The degree of the reciprocal polar cW've is n(n-l)-28.

* ' At a tacnode, two distinct branches of a curve touch, andhave a common tangent and a common point: Therefore, in thereciprocal curve there are two branches having a common tangent at acommon point, i.e., to (t tac'/iode corresponds a tacllode on the l'eciptocatCU1't'e.


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~;~VELOPES 145121. THEOREM: 'P I / I J degree of tlie reciproca! If a ClUTe

with K cusps is n(ii-l)-3K.When the curve has a cusp, the first polar of any point

not only passes through this cusp, but, also has its tangentthe same as the cuspidal tangent (85). Therefore, thecusp counts as three among the intersections of the curvewith its first polar. Consequently, the number of intersec-tions is diminished by three for every cusp on the curve,and this diminution amounts to 3K for the K cusps on thecurve. Hence, tlie degree vf the reciprocai curve 18n(n-l)-!k

Combining this with the preceding' theorem, we obtainthe following :-

TI.,~ dl.'yn!1' < 1 ' flu' reciprocal. polar of a curt irrtl, 8 J1(1d(',~awlKC1f8P8 /" n(it-l)-2o-:1K.

122, 'l'HI.;QRE)!: If a (,W'CI.' IW8 a nuiltip! poiut of order k,the degree of its reciprocal polar eliI've is n(n-l)-k(k-l).

'Ve have seen that [L multiple point of order J . . : on acurve is a multiple point of order k-l 011 the first polar.(86). Therefore, the multiple point counts as k(k-l),~ulOllg the intersections of the cur-ve with its first polar,Consequently, the number of remaining' intersections isn(n-l)-k(k-l), icliict, i8 th e degree of the reciprocal polar('//lTC.""

128. ESYELOl'ES:If the equation of a curve involves a variable

pararneter, we obtain tL series of different curves by givingdifferent va-lues to the parameter. All these curves toucha certain CUl'I'e, which is cn.lled the eucelope of the system,

liA multiple point of order k is equivalent to tk(k-l) nodes,and for each node the class of a curve is diminished by two. Hencefor a multiple point of order k, the class is diminished by k(k-l),which :1grees with the result we have established just now.

19


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1 - 1 : ( ;

I':a \'h


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If the curve f (./',, z, > C ) =0 passes through a tixed point(x ', y ', z'), we have f (.v', y ', z', > C ) =0, and jf this equation issolved for > C , there are found n values of > C , corresponding toeach of which there is a cur-ve of the system, and conse-quently, 'Ib curves of the system pass through any point.When, however, the point lies 011 the envelope, two of thesecurves coincide.E, l. The system uf circles (,v-A)' + u ? = "," where" is a

parameter, has it" centre on the axis of x. The envelope of ihiw.yatem is given by 2 (",-,,)=0, i.e., :v=", which gives for the envelopey : =,.2 , U 1 ', If = ,., i.e., two lines parallel to the locus of centres,E. 2, Tho equation uf the normul at allY point ( ll cos 0, b sin 0) 011the ellipse

is ,,~_...!!J...-=


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1 48 rnsonx oi' PLA.NE CURVESE. -! . To tind the envelope of the chords of curvature of tho

points on 11 conic.The equation of the chord of curvature of the conic . v " i n ' + u ' ; ' " =1

is~ cos a_1( sin a= cos 2aa, Ii

and therefore, the em-elope becomes-

E 5. Find the envelope of a circle of constant radius, whose centremoves on a given conic.

Let a"ja ' + y2.'b' = be the given conic, and (,v-a)' + (!I-/3)2 =.2be the circle, the centre (a , /3) moving on the conic,

Then we may write a=a cos 9, /3=1> sin 9,The equation of the moving circle now becomes--(a'-b') cos 20-'!n,v cos 6-4/J11 sin 6+2(x' +y')+a' +b'-21'2=O.The envelope can now be found as already explained, and is called

the curve prtmlld to the couic.

124. It often happens that the equation of a curvecontains two or more parameters connected by an equationor equations, so as to represent a single variable parameter.In order to determine the envelope we make use of themethod of indeterminate multiplier, the principles * ofwhich we shall presently explain.

Let the equation f ( , I ! . y, Z, A, p.)=0 of a curve contain twoparameters A,p.connected by a relation 1 > (A, p.)=0,

Then we have

It The subject properly belongs to the province of DifferentialGeometry, and the student is referred to Edwards's Differential Calc.Chapter XI, for fuller treatment of the subject.

See also Cayley, Collected Works.


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ENV\';LOPES 149

Eliminating ~ietween these two eqnations, we ohtaill-tf!= a f a f 1 = 0( f X a P - i

I a c p Ia c p lI a X a p - IThe envelope is obtained by eliminating Aand i J . between

the equations f= O , cp=O and t f !=0 .We may eliminate one of the parameters between the

given equations, and then obtain the envelope by the methodof the preceding article.A similar process is followed in the general case, when

the k parameters in the equation of a curve are connectedby k-l relations. In this case the envelope is obtained byeliminating the l: parameters and 1


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I ;j L

1:!;). \\'e have seen tJmt ,i"lU : '1 I the equation of a curveeontains A in the nth degree,i.e., whenwhere a, b, c, .,. are functions of the co-oi-dinates, theanvelope is obtained by equating tu zero the discrirniuant *of f. Thus, for II=2, H , etc., we have for the envelopes

ac=b? =0a+d? +4(lc~+4b3d-6abcd-3b'c' =0, etc.

It will be noticed thai the degree of the envelope IS2(n-1) in the co-efficients a, b, c...Hence, when a, b, (,,,. are lineal'. the degree of the

envelope is 2':U-1).The simplest of these cases is the envelope of

AA' +2BA+C=Owhere A, B, C are all linear functions of the varinbles, so thatthe equation represents a right line.

Eliminating A between

we obtain for the envelope B"=AC, which is a conic.When A, B, C are expressions of the second degree, tIle

envelope B' =AC is a curve of the fonrth degree, and so OIl. It has been proved by Prof. Cayley that the discriminant, in

ceneral, contains other loci bes ides the envelope. In fact the complete, envelope of the variable curve consists of the propel' envelope na~ explained above together with the locus of the nodes of the variable curve

twice, the locus of the cusps thrice, the envelope of the double tangentstwice and the envelope of the stationary tangents tl . ru:.

... Cayley, Messenger of Mathematics, YolsLf and XII. See ulso. , : B.enl'iehi, Proc, of the London Math. Soe., Vol. I r , J. M. Hill, tu,l:l~, and '.Im,", R. ,. Curves, 180 (").


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lil2 1'H~;On.Y Or-PT,AN~ CURVES0I"c/"I('i8~: The equation AA'+2DA+C=O way he

regarded as a quadratic equation to find the values of A forthe two particular members of the family, which passthrough a given point (.t', y). When (:r, y) is a point on theenvelope, these two members must coincide, and conse-quently, the quadratic in Amust have two equal roots, andthe locus of such points is thei-efore B"=AC.

il.x'j+y2!b2=1For, the equation can be put into the form-

.V +!J tan 8= ,la' + b' tan'or, cleariug the radical, ,r' +u" tall' 8+2.1'1/ tau 8=0' + b' tan' 8i.e., (y'-b') tan' 8+2,,y tau O+(.c'-a2)=O.

'I'his, as a quadratic ill tau 0, has two equal roots,if .!.v'y'=4(.e'-a')(y'-b'), i.c., ,V''+'y' b'=1.

E. 2. Show that the envelope of the lilies

,\:cos ma+y Sill /IIa.-=a(cos 11 )-:: -

where" is the arbitrary para ureter, i~ the curve

=(1 cos( . - . ! . . ' . . . . .- ) 8 .JU-n

F ; J . ' . a. Circles are described hav iug' for diameters the radii vectoresfrom the origill to the curve ,va + y'=a" .r '', Prove that their envelopeis the inverse of a semicubioal parabola (Oxford, HiS.'.

E .. 4. An equilateral triangle moves so that two of its sides palilithrough two fixed points P and Q. Prove that the envelope of the thirdside is a circle.

Ex. 5. F'ind the relation between "and b, when the envelope ofthe line ,v.'n+y!b=l is the curve .cPy(=P+'.


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EVOLUTES 153126. EVOLUTES:The evolute of a curve may be defined as (i) the envelope

of its normals, (ii) the locus of the intersection of consecu-tive normals, (iii) the locus of the centre of curvature ateach point of the curve.

It is clearly seen that these definitions can be deducedone from the other.

In order to obtain the evolute of a curve, we generallytake the equation of the normal at any point of the curveand then find its envelope. The same may be found byfinding the co-ordinates of the centre of curvature at anypoint, and then determining the locus of the centre ofcur vature.

The methods will be best illustrated by means of thefollowing simple examples :

Ex. 1. Find the evolute of the parabola y' =4ax.The normal at any point (at', 2at) is-

ta l + y=2at + at" .. (I)To obtain its envelope, we differentiate it with respect to t, and thus get

x=3at' +2a .... (2)Eliminating t bet~een (I) and (2), we obtain

.27ay' =4(x-2a)3

8S the equation of the evolute of the parabola.Ex. 2. Find the evolnte of the ellipse x"/a' + Y'/bl =1.The normal at any point (a cos 8, b sin 8) is-

~-~=a'-b'cos8 sin 8The evolu~eof the ellipse, by Ex. 2, 123, is, therefore,

2 2 ~(ax + (by = (a' -b')',~o


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154 THEORY or PLANg CURVESE . " . 3. Find the evolute of tho oissoid (;c"+ y").,,-(/,y"Writing the equation in the form (n-x)y' = ."3, we may express the

co-ordinates of any point on it in the form((.

,V= (1+82)'The equation of the normal is, therefore,

a(I+202)293 .1 : + (1 + 392)y =--'---:9:---i.e ., 29'z + 39' y-20"", + 9y-a=0.

The discriminant of this is found to contain ('"+ ~a)' + y' as a factor,the remaining factor, therefore, gives the proper evolute, namely,

+ 32 2 0 + 512 3 0Y :3 ((.y- 27'0, ,?)= EJJ. 4. To find the avolute of the curve given by.v-ct, !/=c/t.The centre of curvature at any point (t) is

x=c(3t' + 1)/2t', y=c(t' + 3)/2t.Eliminating t between these, we obtain the equation of the evolute.

127. NOR~IAL O~' THE EVOLUTE :The following construction for the normal of the evolute

is useful:Let I and J be any two finite points in the plane. Then,if the tangent at any point P of a curve meets IJ in M, and

if M' be the harmonic conjugate of M with respect to I, J,then the line PM' may be regarded as the normal. Fromthis it follows at once that if the point P be on the line IJ,then PM' will coincide with that line. But when P coincideswith either I or J, the points M, M' coincide, and the normalcoincides with the tangent.

Therefore, when I and J are circular points at infinity,and P is a point on IJ, the normal at P coincides with theline at infinity, and if the curve passes through either ofthe circular points at infinity, the normal coincides with thetangent.


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l ! l V O L U l ' E S 1 55

128. When a curve is defined by its tangential equation,the line co-ordinates of its normal, and consequently thetangential equation of the evolute, can be easily obtained.

Let c J > a , ' Y J , ,)=0 be the tangential equation of a curve.Then, if a ', ' Y J ' , " ) be the line co-ordinates of any tangent,

(1)

is the equation of its point of contact.Let ",=0 be the equation of a pair of points I, J, then

(2)

is the equation of the pole of the given tangent with respectto IJ, i.e., if P is the point where the tangent meets IJ, thenthe harmonic conjugate Q of P with respect to I, .J is givenby (2). If now I, J are the circular points at infinity,equation (2) gives the point at infinity on the normal.

Therefore the two equations (1) and (2) determine theline co-ordinates of the normal. If, therefore, 1;',' 1 ' / . " beeliminated between the equations (1) and (2), and theequation of the curve, we obtain the tangential equation ofthe evolute.

The equation of the circular points at infinity IS~" + 'YJ"=O .

Then

gives the condition of perpendicularity ~I;'+'YJ'YJ'=O.

Ex. 1. Find the evolute of a central conic.The tangential equation of such a conic is n2e + b ''1/2 =1.. The co-ordinates of the normal are determined by


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156 THEORY 0]


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CAUSTICS 157'Gergonne * has proved that each Katacaustic is the

evolute of an algebraic curve which, according to Quetelet tis called the Secondary Caastic or Anticaustic.

Quetelet has given a practical method by which causticsmay be regarded as evolutes, and in fact, he gives the follow-ing construction :

If with each point successively of the reflecting curve ascentre, and its distance from the radiant point as radius, wedescribe a series of circles, the envelope of all these circleswill be a curve, the evolute of which is the katacansticrequired.

In a like manner, Quetelet has given the followingtheorem:

If with each point successively of the refracting curve ascentre, and a length in a constant ratio to its distance fromthe radiant point as radius, we describe a series of circles,the .envelope of all these circles will be a curve whoseevolute is the Diacansbic.

130. EQUATION OF KATACAUSTICS :Let F(z, y) =0 be the equation of the reflecting curve,

and P(;c', y ') be a luminous point in its plane.Let Q(a, (3) be the point of incidence of a rayon the

curve F. If then T=O and N=0 be the equations of thetangent and normal to the curve at Q, the equation of theincident ray PQ may be written as T+'\N=O, which issatisfied by (x', y'), i.e., PQ is the line TK'-T'N=O, whereT' and N' are the results of substituting x', y' for a : : and yrespectively in T and N,

Now, the incident ray and the reflected ray are equallyinclined to the normal. Therefore, the reflected ray is the

., Gergonne, AnD. de Math., t, 15 (1825), p. 345,t Quetelet, Brux, Ac. Nouv. Mem., Vol. 5 (1829), Nr. 1.


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158 TH]


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CAUSTICS J59and the normal is the line

xsin6-ycos6=O (2)The equation of the reflected ray is, therefore, by thepreceding' article,

(a cos 6+f3 sin 6-r)(x sin 6-y cos 8)+(a sin 6-f3 cos 6)(x cos 8+y sin 6-r)=O

which may be written in the form-

(ay+f3x) cos 26+(f3y-a.~) sin 26+r(;e+a) sin 6-r(y+f3) cos 6=0 (3)

where 8 is a variable parameter.The envelope of this, by the usual method, is found to be[4(a' +f3)(.x +y' )-1" {(x+a)" + (y+ f3)t} ] s

=27(f3.~-ay)(:r.+y-a-f3) (4)which is the equation of the caustic by reflection of a circleand was first obtained by St. Laurent.

1, however, the axis of : e passes through the radiantpoint, and the radius of the circle be taken equal to unity,we have f3=0 and r=l, and the equation of the causticreduces to

{(4at -1)(:1:' +y') -2ax-a'} ~=27a'y'(x' +Y' -a')' (5)

The equation (3) of the reflected ray reduces to the form

( 2 8+ I) + acos26-cos6 + 0- a cos r y a=sin6 . (6)

Differentiating this with respect to 8, we obtaiu

( 2 . 8) + -acos6(l+2sint8)+1 -0- csm X sin'6 -y-


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160 'l'HEORY OF PLANE CURVES

whence, ,1'= u2cosO(I+2sin'0)-uI-3a cos20+2a'

i.e., the co-ordinates (.r, y) of any point on the caustic areexpressed in terms of the angle 0, which is the parameter ofthe point of incidence.

132. TANGENTIALQUATIONF THECAUSTIC:If the equation (6) of the reflected ray be put into the

form ~,c+'YJY+a=O, then we must have-~=-2a cos 0+1acos20-cosO'YJ= sinO

(~-1)~-4a' = -4a2sin'01~'+'YJ'= -.- (I-2a cas O+a')sm'O

andTherefore,

~+al =I-2a cos 0+a2

i.e.,a ' +'YJ'){(~-I)'-4a2} +4a'~+4a4 =0

{~(~-I)-2a'}' +'YJ'{(~_I)2 -4a'} =0 (7)which may be considered as the tangential equation of theCaustic by reflection of a circle.

If, however, (~, 'YJ)be considered as the co-ordinates of apoint, then the equation (7) may be regarded as the equationof the reciprocal polar of the caustic ( 114).

133. If we put y =0 in the equation (5) of the Caustic,we obtain

{( 4a' -I).c -2a;c-a'} s =0


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CAUSTICS 161i.e., -aJ= -- ,20.+1 a:l;=2~1or, the Caustic meets the axis of.u in two points each ofwhich is a triple point.

1 , again, we put Xl +y' =0.', we obtain

and, therefore,

{(40."-1)0.' -20.,'1:-0.' P =0x=-a(I-2a')y= f2a".vI-al

which gives

or, the Caustic meets the circle c>+y"_o.' =0 in two points,each of which is a triple point.

The nature of the infinite branches can be found byconsidering the highest degree terms, i,e., the terms ofdegree six and five. The two asymptotes are thns found tobe given by-

(4a'-I)} r 3 1y= .vI-a' (80.'+1) < . x -40.':"15

By considering the equation of the reflected ray (6), thetangents parallel and perpendicular to the axis of x canbe found easily.

Thus, the parallel tangents are-y=+ .v4a'-:'i- 20.

and the tangents perpendicular to the axis of x are-20.x=

which are, 'in fact"double tangents of the Caustic.Again, when the radiant point lies on the circle, 0.=1,and the equation of t.he curve reduces to-

{3!1'+ (;v-l h3x+ I)}'=27y' (;u" +y" -1)').

"

2 1


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16~. THEORY OF Pf,ANE CURn;gwhich is divisible by (x-I)'. Removing the factor, theCaustic is a curve of the fourth order, uis.,

In the case of parallel rays, the radiant point is atinfinity, i.e., a= ex ,and the Caustic reduces tv-

(4a:~+4y'_1)3 -27y' =0which is a sextic.

We thus see that the Caustic by reflection of a circle isa curve of order 6, has 4 nodes, 6 cusps (including thecircular points), etc. For detailed investigation-seeProf. Cayley's paper on Caustics-CoH. Works, Vol. II,p.357. Also a Memoir by Rev. Hamnet Holditoh-c-QuarterlyMath. Journal, Vol. I(1857), pp. 93111.

Ell). 1. If the incident rays are parallel, show that the caustic is anepicycloid formed by the rolling of one circle upon another of twice itsradius.E. 2. When the incident rays diverge from a point on the

circumference of the reH.ecting circle, show that the caustic curve is acantiQid, which is formed by the rolliug of ,t circle npon another ofequal radius.

EJJ. 3. Rays diverging from the focus of a parabola are reflectedfrom its evolute. Prove that the secondary caustic is a parabola.

E, 4. Rays parallel to the axis of yare reflected from the curvey= e" show that the caustic is the curve_II x+l+ -(x+l)11/-. e e

E : r : . 5. Obtain the caustic hy reflex ion of an ellipse, the radiantpoint being at the centre.

134. INTERSECTIOl'l" OF THE CAlJilTIC WITH THE REFLECTINGCIRCLE:

If iIi the equation of the caustic, we put x' +y' =1, webave-


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VAIJS1'ICS 1 6 3

t.e., {3a'-I-~a.E)-27a(I-,v)(I-a.)' =0which reduces to the form

(ax-l)'(8a.u-27a +18a"+1)=0,The factor' (a.v- 1)' equated to zero shews that the caustictouches the circle at the poiuts

1..v=-,Q

u=x '/1_1- 'V a~i.e., at the points where the circle is met by the polar ofthe radiant point. 'I'he other factor gives-

27a~-ISa'-1.1 := Sa

It can be very easily shewn that the curve passes throughthe circular points at infinity, which are cusps on the curveand the points where the axis of , e meets the curve arecusps (the axis of x being the tangent) and the two pointsof intersection with the circle ;1)' +y' -as =0 are also cusps,the tangent at each point coinciding' with the tangent of thecircle. There are thus ill all 6 cusps.

135. CAUSTIC BY RBFRACTJON 0]


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164 THEORY or PLANE CURVESLet the refracted ray be

y=x tan .p'+c.Since both the rays meet on the line ,1)=0, we must

have-c=tall .p

and the refracted ray becomes-y-x tall .p'=tan .p

or, ksin.py- x-tall .p=0-\1'1- k' sin.p (1)Differentiating this with respect to .p, and combining thtwo equations, we obtain, after a simple reduction-

s(I-k"sin '.1.)-'k;J;=- 'f'coss.p

where

Hance, eliminating .p, we obtai~

which is the required equation of the Caustic.When k< 1, i.e., refraction takes place into a denser

medium, k " is positive, the caustic is the evolute of a hyper-bola.But when k>l, i.e., refraction takes place in a rarer

medium, k " is negative and the eaustic is the evoluteof an ellipse ( 126, Ex. 2). Thus the hyperbola or theellipse is the Antica.ustic or the Secondary Caustic.


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CAUSTICS 16 :1From what has beeu said in 129 we call easily form the

equation of the secondary caustic as follows ;-The equation of the variable circle may be taken as

x"+(y-tan )'-k'sec'=Othe envelope of which is found to be-

k" (.~. +y' -k') _yO =0i.e.,or,which is the equation of the Secondary Caustic, It will beseen that the radiant point it; a focus of the conic.

136. 'vVe have seeu that the caustic by reflection 01'refraction may be regarded as the evolute of a certainenvelope which is called the secondary caustic. In fact, thereflected or refracted rays are the normals tv a series ofsecondary caustics; anyone of these has the reflected orrefracted rays for normals, and consequently the causticcurve for evolute.

It is usually more convenient to tind a secondary causticin some cases than the caustic itself; for instance, in thepreceding article, the secondary caustic by refraction of astraight line is 301\ ellipse or a hyperbola.

We shall now determine the secondary caustic byrefraction at a circle.

Let x' +y' =r' be the refracting circle, p e a , {3) theradiant point, and a, tJ) any point on the circle, so that

~=r cas 9, tJ=r sin fJ.If p. be the refractive index, the secondary caustic win

he the envelope of the variable circlep.1{(x-r cos fJ)"+(y-r sin 8)"}


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1 6 6 l'ltEOltY OF' PLANE CUR.V~S

Writing the equation in the form-e-p.(x" +y' +r')-(a" +f3"+r')

-2r(p.,v-a)cos 0-2r(p.'y-f3) sinO=Othe envelope becomes-

or,

{p.'(;C'+y" +r' )_(a2 +f3' +r")}"=4r" {(p.:t-a) + (p.y -f3)'}

{ fA"Cx' +y' -r' )-(a' +f3' -r')}'=4rp.{(x-a) + (y-f3)'}

If now the axis of ,v is taken to pass through the radiantpoint, f3=0 and a=a (say), then the equation becomes-

{p.(x +y' -r' )-a' + 1 " p =4r"p. {(x-a)' +y'}which, after simplification, may be put into the form

The above equation of the secondary caustic is evidentlyof the form-

mp+tn'p'=c(say)where p and p i are the distances of the point (e, y) on thelocus from the two fixed poin ts

(;., 0) and (a, 0)

respectively, andThus, the locus of (x, y), i.e., the secondary caustic is the

Oval of Descartes or the Cartesian," of which the two fixedpoints are the foci. Therefore the Caustic by refraction of acircle is the eoolute of a Cartesiun. Oval.

See Wiltiamaou'a Dill. Oalculus, Chap. XX, pp. 375.382.


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PRDAL CURVES 167The cases of parallel rays or of rays proceeding from a

point on the circumference of the circle may be deducedfrom the above result" by assuming a=OC: or a=T respec-tively. For further discussions the student is referred toGeometrical Optice by R. S. Heath, Chapter VI, pp. 99~133,and to Prof. Cayley's Memoir.

E m . 1. Prove that the form of the caustic curve near the cusp i8a. semi-cubical parabola.

E m . 2. Find the caustic by refraotion of a circle, when the inoidentrays are parallel.

E m . 3. Rays diverging from the centre of a given circle are reofracted at a curve so that the refracted rays are all tangents to theoircle. Find tbe equation to the refracting curve.

Em . 4. Show that the caustic by refraction of a circle wheu theradiant point is on the circumference is also the caustic by reflectionfor the same radiant point and for a reflecting circle concentric 'withthe refracting circle.

E, 5. Prove that the caust.ic by reflexion of a circle is the evoluteof the limat;on.

137. PEDAL CURVES;The locus of the foot of the perpendicular drawn from

any origin 0 on to the tangent at any point of a curve iscalled the first positive pedal of the curve with respect tothe origin.

The pedal of the first positive pedal is called the secondpositive pedal, the pedal of this latter is called the thirdpositive pedal, and so on.The curve which has the original curve for its firstpositive pedal ifl called the first negative pedal, and so OIl.

Let oy be t.he perpendicular on the tangent at, anypoint P of a curve, and OZ be the perpendicular drawn fromo on to the tangent, at Y to the locus of Y ; then the ang-IeOPY =angle OYZ, or, in other words ;-

The angles between the radiu -cecior asul the tangent atcorresponding points of a curre and it, pedal are equal.


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168 tilEORY 0];' PLANE CURVESIn the same case, we have OP,OZ=OY~. Let OY and

OY' be the perpendiculars drawn from 0 on to the tangentsat two consecutive points P and P' of the curve, the tangeutsmeeting at the point T.

It is clear then, since the angle YOY'= YTY', the points0, Y, Y', Tare concyclic, and therefore,

OYZ=ll'-OYY'=OTY'.Hence, the triangles are similar, and we have-

OZ OY'OY - OT (1)

In the limit when P and P' coincide, the angle OTY'becomes equal to the angle OPY, and OY'=OY, OT=OP.

Hence, OPY=OYZ andOP.OZ=OY, and thetheorem is proved.

In the limit when Pand P' coincide, the circlethrough OYY'T has OP asdiameter and touches thetangent YY' to the pedal.Hence we obtain the

theorem that the circle onradius vector as diameter touches the pedal.

138. THE CARTESIAN EQUATION OF THE PEDAL:Let < p a , 1'/)=0 be the tangential equation of a curve.

Let any tangent to this curve cut the axes of ; v and y in Pand Q respectively.

Let (w , y) be the co.ordinates of Y, the foot of theperpendicular from 0 on to AB.


35/46

UDAL CURVJj l$ 169If L POY=a, we have-

OY=OP co s a = OQsin a,But, OP=I/e .. OQ=I/l];

CDS a sin aIt T---;J 1--~-

Then x=OY cos a=OP cas' ay=OY sin a=OQ sin' a

i.e., and

Hence, ille= . - - - - + . 'x ySubstituting these in the equation of the curve, the loous ofY becomes-

(X y )p - , - =0ill'+y' x' +yll , (1)The inverse of (1) is evidently c p ( ~ , y)=O, which is the

polar reciprocal of the curve ( 114).Hence, we have the definition :-The polar reciprocal of a curve is the inverse of the first

positive pedal, and the pedal is the inverse of the polarreciprocal curve.

Oor.: The tangential equation of the first negative pedalis-

when c p ( c c , y)=O is the point-equation of the curve,2~


36/46

170 THEORY OF PLANE CURV~sEx. 1. The first positive pedal of the curve (x/a)' + (y/b)''''l is

n (x


37/46

PAR..1LI.EL CURVES Inare evidently given by ttn =0, which shows that the tangentsu.=0 are parallel to the asymptotes of the original curve.

The equation (2) shows t.hat the inverse curve has alsoa.multiple point of order n at each of the circular points atinfinity.

The following examples will illustrate some of theproperties of inverse curves, and can easily be solved withthe help of the formulae established in 15.

EJ) . 1. Find the inverse of a conic 1V. r. t. IIfocus as the pole.EJ) . 2. Prove that the inverse of the curve ap, + (JP. + 'YP. =0 tv. r. t.

any origin is a curve whose equation is of the same form.EJ). 3. Show that to a double point on auy curve corresponds

another double point of the same kind on the inverse curve with respectto any origin.

EJ). 4. Show that the inverse to the kth positive pedal is the k.thnegative pedal of the inverse curve.

EJ). 5. The osculating circle at any point of a cnrve inverts, ingeneral, iuto the osculating circle of the inverse curve at the inversepoint.

E, 6. Discuss the case when the osculating circle passes throughthe origin.

E. 7 . Find the number of osculaHugcircles of IIgiven CurvewhichpaBsthrough a giveu point.

140. PARALLEL CURVES:DEFINITIO~: The envelope of a line parallel to the tangent

of a given curve at a fixed distance is called a curve parallelto the given one.From this defiuition it follows that, if on the normal at

any point P to a.given curve, a point Q is taken such thatPQ=k=const.

the locus of Q is a curve parallel to the given curve. Itfollows then that all parallel curves have the same normals


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172 THEORY OF PLANE CURVESand the same evolute; but every normal to a parallel curveId normal in two places corresponding to the values k.

The two possible positions of Q corresponding to thevalues k are, in general, branches of the same curve. Butthey may be different curves in certain cases.

Letbe the equation of a tangent to a curve. Then, the equationof a parallel line at a distance k from it may be written as

the envelope of which will be a parallel curve, If thencpa, Y J , ') =0 be the tangential equation of a curve, that ofthe parallel curve is obtained by writing

for' in the given equation, i.e., the equation of the parallelcurve is-

A parallel curve may also be regarded as the envelopeof a circle of given radius whose centre moves along thecurve. The methods will be clearly illustrated by meansof a few examples.

E x , L b'illd the parallel to the curve+-

'I'he cu.ordiua.t.es of an~' point 011 thi~ curve may be takea asa (;08'8, (1 . ~in"8.

The equa.tioll of the tallg"llt is theur ':08 8 + y sill 8=(1. Sill 8 cos II.

no th~.r, of a, Ii ne paralle I to this, at a distance k, Is-: t : 0011~+tI.il1 '=k+C! sin 8 COI ,.


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rSOPTlC LOCI 173The envelope of this line is found to be-

{3(x" + y' - a") -4k'} 3 + {27a))y-9k(,o' + y') -18a'k + 8k3 } 2 =0.In this case there are two different curves constituting the parallel

and these are not branches of the same curve.The tangential equation is found to be-

E.,;. 2. In the case of a circle of radius /', the parallel becomes twoconcentric circles, of radii I' k, and these are certainly different curves.

Ex. 3. To find the tangential equation of the parallel to

The tangential equation uf the ellipse is (I'e + b ' , , ' =(', and con-sequently, that of the parallel is-

a 2 e +b''''=(+kV~2 +'7')'or {(a2 - /,;2)e + (b' _ /,;'),,2 _ (2} 2=4k2(~2 + ,,2)('.

Ex. 4. Show that th tangential equation of the parallel to theparabola y'=4a,1J is-

E,u.6.constant.

'l'he radii of a co-axial system of circles are increased by BShow that their new f'nvelope i . q parallel to their old.

141. ISOPTlC LOCI;The locus of the point of intersection of two tangents

which cut one another at a given angle is called an IsopticLocus of the given curve. If the tangents cutting at anaIlgle 7"-U are included among those cutting at the angle(I., the Ioeus is an algebraic curve.

The method o f finding tlH' lo e lll-; wi ll h


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174 THEORY OF PLANE CURVESis a tangent to the parabola. Hence the slopes of the two tangentspassing through any point (",', V') are given by

m2.v'-mv' +a=O.If m, and 1 . are the slopes. then

and

V y " I x " - 40 l x 'l+~~'

V.y" - 4ax'a+ro'

:. '1he locus of (x'. y ') becomes-v- 4ax= tan 2"(a + x)'.

Eili. 2. Find the isoptic locus for x'fa' +y'fb'=lThe tangential equation of the ellipse is-

(1)If we eliminate ( between this and ~ll+ 'lY= (, the resulting equation is

(2)If now 1I.=tan 8= -~/'l be the slope of the tangent drawn from anypoint (w , V), we obtain from (2) the equation-

ml(a' -ro') + 2wym + (b' - Y')=Owhich gives the slopes of the two tangents.

Now, 'Ill, and nt. being the slopes, we obtain--2wy

'Ill, +tn.= as_x' b'-Vlm 11n3= a:1-a:'

.whenceV4b'0I)'+4a'y'-4a'b'

-tan "= w'+y'-u'-b'i.e., the equation of the locus is-

4(b'z. +a'y'-a'b')=(x' +Y'-a'-b2)' tan '".'The caee of a general n.ic assumes a simpler form, when the

COD~tantankle is a riiht anile, and will be taken up .hortly.,, .!;


41/46

ORTHOPTIC LOCI 17f i

142. If PQ and PR are two tangents to a . curve inclinedat a constant angle, the circle PQR touches the isoptic locus.

be consecutivepoints on th eisoptic locus, andPQ, PR aud P'Q'P'R' be tangentsinclined at anangle a to eachother. Then, Q,P,P', Rare eoncyclic,since L QPR= L QP'R, and in the limit, the circle PQRtouches the locus of P.

It is clear then that if the tangents PQ and PR to acurve are inclined at a constant angle and the normals at Qand R meet at S, then PS is the normal to the isoptic locus.

Let P and P

QQ '

R '

Since QS and RS are perpendicular to the chords PQand PR of the above circle, the point S lies on thiscircle, and PS is a diameter and is consequently perpendicularto the tangent at P to the circle and the isoptio locus, thatis to say, PS is the normal at P to the isoptic locus.

It can be easily shown that the isoptic locus hasmerit-I)-pIe points at I and J.

143. ORTHOPTIC LOCI: ' * .The locus of the point of intersection of two tangents

to a curve which cut one another at right angles is calledthe orthoptio locus of the curve.

This is a particular class of isoptic locus, when theconstant angle is a right angle.

* Dr. C. Taylor-U Note of a Theory of Orthoptic and IsopticLocus," Proc, of the Royal Soc. London, Vol. 37 (1884), pp. 138141.


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.176 TH~ORY OF PLANE CURVES

CAI'Tb:~IAN BQliATIO:O;:Let cp(~, 1'])=0 be the talJgential equation of the curve,

and ~.r.+ 1']Y +1=0, that of a point P on the locus, so that twoof the tall gents drawn from P to the curve are perpendicularIf now we make cpa, 1 ']) =0 homogeneous in ~, 1 '] by meanso f ~.t:+'l}y+l=O, the resulting equation in ~/1']gives the slopeo f the tangents which can be drawn from P to the curve.

Putting -~/1']=m. the resulting equation can be put intothe form l/ I;rn)=O (1)

If two tangents are at right angles, two of the rootsm" m. of (I) mnst be connected toget.11er by the equntion

n~lml +1=0,The condition 01' this is that the eliminant of

should vanish, which gives a relation between . 1 : and y ,which is the orthoptic locus.Since this eliminant for it curve of the mth class is of

degree (m-I) in the co-efficients, which are themselves, ingeneral, of degree rn, the degree of the orthoptic locus of acurve of the mtll class cannot be greater than m(m-I).Ex. 1. Find the orthoptic locus of the parabola,

This follows from . E r e . 1 141,when" .:=-.2

By using the method of the present article, we proceed 1 1 . I follows:-The tangential equation of the parabola i~ ((1/" =~.~{akingt.his homogeneous,we get

Putting -~/1/=m, weobtain m':n-my+a=OPutting m= -m-l, we obtain-

.. (1)

(2)


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ORTHOPTIC LOCI 1 7 1

From (I) and (2), we have-

whence {(a -x)' +y' }(a: + a)' =0which gives a: + a =0 as the locus required.

Ex. 2. The orthoptic locus ofx 'la ' + y' Ib ' =1

is found to be x' + y' =a' + b" which is the director circle.Ex. 3. In the case of the circle, x' +y'=I", it becomes a:' +y'=2r'which is a concentric circle.E, 4. Find the orthoptic locus of the evolute of the parabola

y'=4ax.The tangential equation of the evolute is 4ap = 27'11'Making this homogeneous, we obtain-

Putting -~/'1=m, we get40m3 + 27;vm-27y=O ... (I)

Putting m= _m-1 in (1) we obtain27ym3 + 27ilJm' + 4a=O ... (2)

Now, the eliminant of (1) and (2) will give the required orthoptic10CllS.

E. 5. The orthoptic locus of the evolnte of fI)'/a' +y'jb'=1is (a' + bS)(x' + y' )(a'y' + b' fI)')' = (as - b')'(a'Ys - b'x')'a sextic curve.

144. If the equation of the curve be given m theparametric form X=il(t), y=i.(t), we may proceed to findthe orthoptic locus as follows:

Let t and t' be the parameters for any two points Pand Q on the curve.


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178 'l'HIWRY OJ-' PLAN"; CURVES

Then the equations of the tangents at P and Q may bewritten in the forms-x+cf>(t)y=tf!( t ) and x+cf>(t ' )y=tf!( t ' ) (1)

where cpand of!are functions of t.The slopes of these tangents are-

and 1cf>(t ' ) "Hence, if they are perpendicular, we must have

cp(t). cp (t')+ I=O (2)If now we make the substitution u=t+t' and v=tt',

then, by means of equations (1) and (2) we can express (, y)the co-ordinates of the point of intersection of (I) in termsof a.parameter u or v.

Ex. 1. Consider the curve x=at2, y=2atTangents at two points whose parameters are t and t' are

w-t'y+at" =0Since they are perpendicular, tt' + 1=0, whence we get

-y--a(t" -1) 1t'i.e., W + a= 0 is the orthoptic locus.Ell. 2. Consider the parabola-

. 1 )=01" + 2bt + C 2 < p ( t ) (say)y=At' + 2Bt +C-= .oHI ) (say)

The equation of the tangent at any point is-w l j l ' ( t) - y < p ' ( t) = < p (tW(t) -


45/46

Olt1'j-lOl"1'1C LOCI 179Here, .p(t)=at' +2bl +c

Similarly,'(t)=2at+2bf(t) =2At + 2B,

:. The equation of the tangent is-

~(2At + 2B) -y(2al + 2b)= (Ab-aB)t' + (Ca -cA)t + (Bc-Cb).=J(I) (say) (1)

Similarly, tho taugeut at another point (I') iij:U(2At ' + 2B)-1I(2((t' +2b)= i"(t'). .., (2)

Since the tangents are perpendicular, we hlLVc-

' t J . . 1 . >/IJ!} + 1=U< /> '(1) < /> '(1 ') ... (3)

If now we put It=l+t', u=Lt", then, by means of equations (1), (2)and (3) we can find x and 1 1 iu terms of u or v.

145. EQI:.\T10~ or THE ouruorrrc LOC\.:S WHE)f TH; POLAREQUATIO)f OF A CURVE IS GIVE)f:

Let l'=f(8) (1)be the polar equation of a curve.of its first positive pedal canthe form

Then the polar equationbe obtained, as usual, in

(2)In this latter equation r is evidently the length of the

perpendicular drawn from the pole on the tangent, and a isthe angle which this perpendicular makes with the initialline.

Writing the equation of a tangent in the form-e-x cos a+y sin a=p,

we have from equation (2)(3)


46/46

180 THEORY 0],' PLANE CURV.I!lS

A perpendicular tangent makes with the initial line anangle

(2k+l); +aand consequently, its equation can be written as-

-,I;sin a+y cos a=(-1)k{(2k+l)~ +a} (4)

From equations (3) and (4) we can express J: and y interms of a.

E:e. 1. Find the orthoptic locus of the curve ""=u" cos m8.The equation of the pedal is-

l' m:l =a m~l COS ~ 8.m+ 1

orIH+l

"=u f cos _'_"_ 81 -.,-t m+l jHence, the equation of any tangent may be written as-

{ )~Q) cos a+y sin a=u oos ~ 8 l *m+ 1 )whence, proceeding as in the article, the equation of the orthopticlocus can be found,

Theorems of 142also hold for the orthoptic loci.EiKJ. 2. Find the orthoptic locus of the Cardioid I '=a( l + cos 8),

[The locus consists of a circle and a Iimacon.]

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Ganguli, Theory of Plane Curves, Chapter 6

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