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S. L. (Sidney Luxton) Loney.

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37. If a rectangular hyperbola circumscribe a triangle, shew that
it meets the circle circumscribing the triangle in a fourth point, which
is at the other end of the diameter of the hyperbola which passes
through the orthocentre of the triangle.

Hence prove that the locus of the centre of a rectangular hyper-
bola which circumscribes a triangle is the nine-point circle of the
triangle.

38. Two rectangular hyperbolas are such that the asymptotes of
one are parallel to the axes of the other and the centre of each lies on
the other. If any circle through the centre of one cut the other again
in the points P, Q, and jR, prove that PQR is a triangle such that each
side is the polar of the opposite vertex with respect to the first
hyperbola.



20



CHAPTER XIV.



POLAR EQUATION OF A CONIC SECTION, ITS FOCUS
BEING THE POLE.



335. Let S be the focus, A the vertex, and ZM the
directrix ; draw SZ perpendicular to ZM.

Let ZS be chosen as the positive direction of the
initial line, and produce it to X,

Take any point P on the
curve, and let its polar co-
ordinates be r and 6, so that
we have

SP = r, and iXSP^O.

Draw PN perpendicular
to the initial line, and PM
perpendicular to the directrix.

Let SL be the semi-latus-
rectum, and let SL = I.

Since SL = e , SZ^ we have

szJ-,

e
Hence
r = SP = e.PM=e,ZN

= e{ZS + SN)
I




= ei- + SP. cosO



Therefore



r =



= l + e.r. cos 0.
1



1 — 6 008 6^



.(1).



THE POLAR EQUATION, FOCUS BEING POLE. 307

This, being the relation holding between the polar
coordinates of any point on the curve, is, by Art. 42, the
required polar equation.

Cor. If SZ be taken as the positive direction of the initial line and
the vectorial angle measured clockwise, the equation to the curve is

I

r= .

1 + e cos d

336. If the conic be a parabola, we have e = l, and the equation

I I I ^e

IS r = z = — ^— = - cosec^ - .

1-cos^ ^ . ^d 2 2

2 sm^ -

If the initial line, instead of being the axis, be such that the axis
is inclined at an angle y to it, then, in the previous article, instead of
6 we must substitute d -y.

The equation in this case is then

-=l-ecos(^-7).

337. To trace the curve — = \—e cos 6.

r

Case I. e = 1^ so that the equation is - = 1 - cos 6.

When is zero, we have - — 0, so that r is infinite. As

r

increases from 0° to 90°, cos^ decreases from 1 to 0,

and hence - increases from to 1, i.e. r decreases from
r

infinity to I.

As increases from 90° to 180°, cos^ decreases from

to — 1, and hence - increases from 1 to 2, i.e. r decreases
r

from I to ^l.

Similarly, as changes from 180° to 270°, r increases

from - to I, and, as changes from 270° to 360°, r increases
from ^ to CO .

The curve is thus the parabola co FPLAL'P'F' oo of
Art. 197.

20—2



308 COORDINATE GEOMETRY.

Case II. e<l. When is zero, we have - = 1— e,

r

i.e. r— r . This gives the point A' in the figure of Art.

247.

As 6 increases from 0° to 90°, cos B decreases from 1 to

0, and therefore 1— ecos^ increases from 1 -e to 1, i.e. -

r

increases from 1— e to 1, i.e. r decreases from to I.

1— e

"We thus obtain the portion A'PBL.

As 6 increases from 90° to 180°, cos^ decreases from
to — 1, and therefore 1—e cos increases from 1 to 1 + e,

i.e. - increases from 1 to 1 + e, i.e.r decreases from I to r .

r 1 +e

We thus obtain the portion LA of the curve, where

SA = ^.

Similarly, as increases from 180° to 270° and then to
360°, we have the portions AL' and L'B'P'A'.

Since cos 6 = cos (— 6) = cos (360° — 6), the curve is sym-
metrical about the line SA'.

Case III. e > 1. When is zero, 1 -e cos 6 is equal
to 1—e, i.e. — (e— 1), and is therefore a negative quantity,
since e > 1. This zero value of gives r = — I — (e — 1).

We thus have the point A' in the figure of Art. 295.

Let 6 increase from 0° to cos~^ ( " ) • Thus 1—e cos
increases algebraically from — (e — 1) to — 0,
i.e. — increases algebraically from — (e — 1) to — 0,

i.e. r decreases algebraically from = to — cc .

For these values of 6 the radius vector is therefore
negative and increases in numerical length from to oo .



THE POLAR EQUATION, FOCUS BEING POLE. 309

We thus have the portion A'P^R oo of the curve. For
this portion r is negative.

If 6 be very slightly greater than cos~^ - , then cos 6 is

slightly less than - , so that 1 -e cos is small and positive,

and therefore r is very great and is positive. Hence, as 6

increases through the angle cos~^ - , the value of r changes

from — cx) to + 00 .

As 6 increases from cos~^— to tt, 1— ecos^ increases

e

from to 1 + e and hence r decreases from oo to q .

1 +e

Now is < — - . Hence the point A, which corresponds

to 6^ = TT, is such that SA < SA'.

For values of 9 between cos~^- and tt we therefore

e

have the portion, ao RPA^ of the curve. For this portion
r is positive.

As increases from tt to 27r — cos~^ - , e cos 9 increases

e

from — e to 1, so that 1 — ecos^ decreases from 1 +e to 0,

and therefore r increases from — to oo . Corresponding

to these values of 6 we have the portion AL'R^ oo of the
curve, for which r is positive.

Finally, as increases from Stt-cos"^- to 27r, ecos^

e

increases from 1 to e, so that 1 — e cos 6 decreases algebraic-
ally from to 1 — e, i.e. - is negative and increases

r

numerically from to e— 1, and therefore r is negative and

decreases from go to . Corresponding to these values

of 6 we have the portion, oo R{A\ of the curve. For this
portion r is negative.



310 COORDINATE GEOMETRY.

r is therefore always positive for the right-hand branch
of the curve and negative for the left-hand branch.

It will be noted that the curve is described in the order

A'P^R' 00 00 RPAL'R^ oo oo R^A'.

338. In Case III. of the last article, let any straight line be
dravm through S to meet the nearer branch in ^, and the further
branch in q.

The vectorial angle of p is XS-p, and we have

I
^~l-e cos XS;p '

The vectorial angle of q is not XSq but the angle that qS produced
makes with SX, i.e. it is XSq^ir. Also for the point q the radius
vector is negative so that the relation (1) of Art. 335 gives, for the
point a,

^ l-ecos{XSq-f^7r) 1 + ecosXSq'

*-^* ^^^~l + ecosXSq'

This is the relation connecting the distance, Sq, of any point on
the further branch of the hyperbola with the angle XSq that it makes
with the initial line.

339. Equation to the directrices.

Considering the figure of Art. 295, the numerical values

of the distances SZ and SZ' are - and - + 2CZ,

e e

I.e. - and - + 2



e e e(e2-l)'

since ^^=-e-^^)' [Art. 300.]

The equations to the two directrices are therefore

r cos o = — ,

e

. rl 21 -] le^+1

and r cos c^ = — - + — — - — — = — - .

\je e(e^ — 1)J e e^—l

The same equations would be found to hold in the case
of the ellipse.



POLAR EQUATION TO A CONIC. 311

340. Equation to the asymptotes.

The perpendicular distance from S upon an asymptote
(Fig., Art. 315)

= OS sin ACK^ = as . . ^ = h.
s/a" + }p

Also the asymptote CQ makes an angle cos~^ — with the
axis. The perpendicular on it from S therefore makes an
angle ^ + cos-i - .

Hence, by Art. 88, the polar equation to the asymptote
CQ is

h = r cos ^ — 9 — cos~^ - = r sin 6 — cos~^ - .
The polar equation to the other asymptote is similarly
h — r cos 6- y-^ cos "^ - ) = — r sin {d + cos~* - j .

341. Ex. 1. In any conic, prove that

(1) the sum of the reciprocals of the segments of any focal chord
is constant, and

(2) the sum of the reciprocals of two perpendicular focal chords is
constant.

Let PSP' be any focal chord, and let the vectorial angle of P be a,
so that the vectorial angle of P' is ir + a.



(1) By equation (1)


of Art. 335, we have






^p=l ecosa,


and


A =1

SP'


- e cos (tt + a) = 1 + e (


Hence




SP^SP'~ '


so that




112

SP'^ SP'~ I'



The semi-latus-rectum is therefore the harmonic mean between
the segments of any focal chord.



.312 COORDINATE GEOMETRY.

(2) Let QSQ' be the focal chord perpendicular to PSP', so that the
vectorial angles of Q and Q' are - + a and -^ + a. We then have



7^7^ = 1 -e cos ( - + a | = l + esina,






and — - = l-ecos (-|^ + aj = l + ecos(^ + aj=l-esina.

Hence

7 7 27

PF'=SP + SP' = , +



1- e cos a 1 + e cos a 1 - e^ cos^ a '
Z Z 21

^^ ^ ^ l + esma 1-e sin a 1-e^ sm^ a
Therefore

1 1 1-e^ cos^ a 1 - e^ sin^ a _2-e^

PP''^'QQ'^ 2Z "^ 21 ~ ^2F'
and is therefore the same for all such pairs of chords.

Ex. 2. Prove that the locus of the middle jioints of focal chords of
a conic section is a conic section.

Let PSQ be any chord, the angle PSX being d, so that

I



SP

and SQ=

Let .
r and d.



1-e cos d '
I I



1-e cos [it + 6) 1 + e cos '
Let iJ be the middle point ot PQ, and let its polar coordinates be



Then .=SP-BP=SP_«^+^= «^-««



^ Ll-«cos^ 1 + e COS ^J 1-



2

ecos^
e'-^ cos 20'
i.e. r^-eVcos^^rrZe. ?'cos0.

Transforming to Cartesian coordinates this equation becomes

x^ + y^- e^x^ =lex (1) .

If the original conic be a parabola, we have e = l, and equation (1)
becomes y^ = lx, so that the locus is a parabola whose vertex is S and
latus-rectum I.

If e be not equal to unity, equation (1) may be written in the form

^ (i-')[-ir^J+'/=4-(^j

and therefore represents an ellipse or a hyperbola according as the
original conic is an ellipse or a hyperbola.



POLAR EQUATION TO THE TANGENT. 313

342. To find the 'polar equation of the tangent at any
point P of the conic section - — \~e cos B.

Let P be the point (rj, a), and let Q be another point
on the curve, whose coordinates are (r^, /?), so that we have

1 — ecosa (1),



T



and - = 1 - e cos i3 (2).

By Art. 89, the polar equation of the line PQ is
sin(;g-a) _ sin(6>-a) sin (j8 - 6)
r r^ ^1 '

By means of equations (1) and (2) this equation becomes

- sin (^ - a) = sin (^ — a) {1 — e cos /3} + sin {(i-6){\—e cos a}

= {sin(^ -a) + sin {(3-0)} -e {sin (^ -a) cos^ + sin (/3-^) cos a}

^ . B-a 20-a-B
= 2 sm ^— ^r — cos

— e{(sin^ cos a - cos ^ sin a) cos ^ + (sin /5 cos ^— cos ^ sin ^) cos a}



2 sin ^—^ — cos [B ~ J — e cos sin ifi - a),

^^_^A_ecos^ ..,(3).



I B-a //,a + y8

».e. - = sec — - — cos ' "

r A

This is the equation to the straight line joining two
points, P and Q^ on the curve whose vectorial angles, a and
^, are given.

To obtain the equation of the tangent at P we take Q
indefinitely close to P, i.e. we put /5 = a, and the equation
(3) then becomes

- =cos (^ — a) — ecos^ (4).

This is the required equation to the tangent at the
point a.



314 COOKDINATE GEOMETKY.

343. If we assume a suitable form for the equation to the
joining chord we can more easily obtain the required equation.

Let the required equation be

-=Lcos(^-7) -ecos^ (1).

[On transformation to Cartesian coordinates this equation is
easily seen to represent a straight line ; also since it contains two
arbitrary constants, L and 7, it can be made to pass through any two
points.]

If it pass through the point {r^^ a), we have

1 - e cos a =— =Xi cos (a - 7) - e cos a,

i.e. I, cos (a -7) = ! (2).

Similarly, if it pass through the point {r^, /3) on the curve, we have

Lcos(/3-7) = l (3).

Solving these, we have, [since a and ^ are not equal]

Substituting this value m (3), we obtam i = sec— ^ .



The equation (1) is then



(-^0-



-=sec— — COS ( d ^ ) -ecos^.



As in the last article, the equation to the tangent at the point a is
then

I

-=QOS{d-a)-eGosd.
r

^344. To find the polar equation of the polar of any

point (r^ , ^1) with respect to the conic section ~ — \—e cos 0.

Let the tangents at the points v^hose vectorial angles
are a and /? meet in the point (r^, ^1).

The coordinates r^ and 61 must therefore satisfy equation
(4) of Art. 342, so that

— — cos(^i — a)-ecos^i (1).

Similarly,

- = cos{6^- 13) -ecose^ (2).



• POLAR EQUATION TO THE POLAR. 315

Subtracting (2) from (1), we have

cos ($1 — a) = cos (^1 — /?),
and therefore

^1 - a = — (^1 — /?), [since a and /? are not equal],

2
Substituting this value in (1), we have



t.e



= ^1 (3).



— = COS \ — ^r- ay —e cos d-i .

n [ 2 )



%.e. cos ^-^r — = — hecos^, (4).

Also, by equation (3) of Art. 342, the equation of the
line joining the points a and fi is

- + e cos d = sec -— — cos ' "
T 2



( - + e cos ij \ cos — - — = cos (6^ — ^ J ,

(- + ecos^ j f- + ecos^ij = cos(^-^i) (5).



%,e.



This therefore is the required polar equation to the polar
of the point (r^, 6-^.

^345. To find the equation to the normal at the point
whose vectorial angle is a.

The equation to the tangent at the point a is

- = cos (0 — a) — e cos 0,

i.e., in Cartesian coordinates,

a? (cos a — e) + yBina = l (1).

Let the equation to the normal be

^cos^ + ^sin^ = - (2),

i.e. Ax + Bi/ = l (3).



316 COORDINATE GEOMETRY.

Since (1) and (3) are perpendicular, we have

A (cos a-e) + Bs>ma = (4).

Since (2) goes through the point ( ^"^ , a j we have

A cos a + ^ sin a = 1 — e cos a (5).

Solving (4) and (5), we have

1— ecosa , „ (1 — ecosa) (e — cosa)

A = , and B = /^— .

e e sm a

The equation (2) then becomes

Zesina
sm a cos + {e — cos a) sm =



r(l — e cos a) '

e sin a I

I.e. sui(0 — a)~esuid = — ^ — .-.

^ ^ l—e cos a r

346. If the axis of the conic be inclined at an angle y to the
initial line, so that the equation to the conic is

- = 1-6 cos (^-7),
r

the equation to the tangent at the point a is obtained by substituting
a - 7 and ^ - 7 for a and 6 in the equation of Art. 342.

The tangent is therefore

-=e cos {d-a)-e cos [d - 7).

The equation of the line joining the two points a and p is, by the
same article,



I B-a

- = sec^-^r- cos
r 2



The equation to the polar of the point {i\ , 6^) is, by Art. 344,
j- + gcos(^-7)l j- + e cos (^1-7)1 =cos{d-d{).

Also the equation to the normal at the point a

■ ,« V . , ^x-> elsm(a-y)

r {e sm ^ - 7 + sin a -6)} = , , \ •

*- \ 1/ \ 'i l-ecos(a-7)

347. Ex. 1. If the tangents at any two points P and Q of a
conic meet in a point T, and if the straight line PQ meet the directrix
corresponding to S in a point K, then the angle KST is a right angle.



POLAR EQUATION. EXAMPLES. 3l7

If the vectorial angles of P and Q be a and /3 respectively, the
equation to PQ is, by equation (3) of Art. 342,



I ^-

- = sec^
r 2



^cosT^-'^Vgcos^ (1).



Also the equation to the directrix is, by Art. 339,

-= -ecos^ (2).

r

If we solve the equations (1) and (2), we shall obtain the polar
coordinates of K.

But, by subtracting (2) from (1), we have

so that SK bisects the exterior angle between SP and SQ.

Also, by equation (3) of Art. 344, we have the vectorial angle of T

equal to ^— , i.e. L TSX=



2 ' 2

Hence Z KST= L KSX - z TSX='^ .

Ex. 2. S is the focus and P and Q tioo points on a conic such that
the angle PSQ is constant and equal to 25; prove that

(1) the locus of the intersection of tangents at P and Q is a conic
section ivhose focus is S,

and (2) the line PQ always touches a conic ivhose focus is S.

(1) Let the vectorial angles of P and Q be respectively 7 + 5 and
7 - S, where 7 is variable.

By equation (4) of Art. 342, the tangents at P and Q are therefore
- = cos(6*-7-5)-ecos^ (1),

and - = cos(6'-7 + 5)-ecos^ .(2).

If, between these two equations, we eliminate the variable quantity
7, we shall have the locus of the point of intersection of the two
tangents.

Subtracting (2) from (1), we have

cos {e - y-8) = cos (^ - 7 + 8).
Hence, (since 5 is not zero) we have 7=^.



318 COORDINATE GEOMETRY.

Substituting for 7 in (1), we have

-=cos5-gcos^,
r

Isead ^

i e. =1 - esec ocos^.

r

Hence the required locus is a conic whose focus is 8, whose latus
rectum is 21 sec 5, and whose eccentricity is e sec 5.

It is therefore an ellipse, parabola, or hyperbola, according as
e sec 5 is < = >1, i.e. according as cos 8> = <:e.

(2) The equation to PQ is, by equation (3) of Art. 342,

- = sec5cos(^- 7) -ecos^,

T

i,e. =cos(&-7) -ecos dcos^ (3).

T

Comparing this with equation (4) of Art. 342, we see that it always
touches a conic whose latus rectum is 21 cos 5 and whose eccentricity
is ecosS.

Also the directrix is in each case the same as that of the original

^ , ^, Z sec 5 - i cos 5 ^ ^ I

conic. For both r and are equal to - .

e sec 8 e cos 5 e

Ex. 3. A circle passes through the focus S of a conic and meets it
in four points ivhose distances from S are r^, ?2, r^, and r^. Prove that

dH^
(1) r-^r^r^r^ = —^ , lohere 21 and e are the latus rectum and

eccentricity of the conic, and d is the diameter of the circle,

. /«x 1 1 1 1 2

and (2) - + - + - + - = 7.
J-i r^ rg 7-4 I

Take the focus as pole, and the axis of the conic as initial line, so
that its equation is

- = l-ecos^ (1).

If the diameter of the circle, which passes through S, be inclined
at an angle 7 to the axis, its equation is, by Art. 172,

r=dcos{d-y) (2).

If, between (1) and (2), we eliminate 6, we shall have an equation
in r, whose roots are ?i, r^, r^, and r^.

r — l I (r — 1\^
From (1) we have cos 6= , and hence sin ^= \/ ^~\ ) ♦

and then (2) gives

r=d cos 7 cos 9 + d sin 7 sin 6,
i.e. {er^ -dco8y{r - l)}^=d^ sin2 7 [eV _ (^ _ 1^2^^

i.e. eV- 2edcos7 . r3+r2 {d^ + 2eldGosy- e2d2sin27) - 2ld^r+dH^=sO.



POLAR EQUATION. EXAMPLES. 319

Hence, by Art. 2, we have

'V2^3^4 = -^ (%

and »'2''3''4 + r3r^»i + ?4rir2 + rir2r3=-^ (4).

Dividing (4) by (3), we have

11112

-+-+- + - = 7. •
rj ra r. r^ I

EXAMPLES. XXXIX.

1. In a parabola, prove that the length of a focal chord which is
incHned at 30° to the axis is four times the length of the latus-rectum.

The tangents at two points, P and Q, of a conic meet in T, and S
is the focus ; prove that

2. if the conic be a parabola, then ST^=SP . SQ.

3. if the conic be central, then — - -^7^;^ = —^ , where b is
the semi-minor axis.

4. The vectorial angle of T is the semi-sum of the vectorial
angles of P and Q.

Hence, by reference to Art. 338, prove that, if P and Q be on
different branches of a hyperbola, then ST bisects the supplement of
the angle PSQ, and that in other cases, whatever be the conic, ST
bisects the angle PSQ.

5. A straight line drawn through the common focus *S of a
number of conies meets them in the points Pj, P^, ... ; on it is taken
a point Q such that the reciprocal of SQ is equal to the sum of the
reciprocals of /SPj, SP^,.... Prove that the locus of Q is a conic
section whose focus is 0, and shew that the reciprocal of its latus-
rectum is equal to the sum of the reciprocals of the latera recta of the
given conies.

6. Prove that perpendicular focal chords of a rectangular hyper-
bola are equal.

7. PSP' and QSQ' are two perpendicular focal chords of a conic ;
prove that ^^ + ^-^-^-^ is constant.

8. Shew that the length of any focal chord of a conic is a third
proportional to the transverse axis and the diameter parallel to the
chord.

9. If a straight line drawn through the focus S oi a. hyperbola,
parallel to an asymptote, meet the curve in P, prove that SP is one
quarter of the latus rectum.



320 COORDINATE GEOMETRY. tExS.

10 Prove that the equations - = l-ecos6 and -=-ecos0-l

r r

represent the same conic.

11. Two conies have a common focus; prove that two of their
common chords pass through the intersection of their directrices.

12. P is any point on a conic, whose focus is S, and a straight
line is drawn through 5^ at a given angle with SP to meet the tangent
at P in T ; prove that the locus of T is a conic whose focus and
directrix are the same as those of the original conic.

13. If a chord of a conic section subtend a constant angle at the
focus, prove that the locus of the point where it meets the internal
bisector of the angle 2a is the conic section

ZC0S5 ^ r. n

= l-e cos 8 COS 6.

r

14. Two conic sections have a common focus about which one of
them is turned ; prove that the common chord is always a tangent to
another conic, having the same focus, and whose eccentricity is the
ratio of the eccentricities of the given conies.

15. Two ellipses have a common focus ; two radii vectores, one to
each ellipse, are drawn from the focus at right angles to one another
and tangents are drawn at their extremities ; prove that these tangents
meet on a fixed conic, and find when it is a parabola.

16. Prove that the sum of the distances from the focus of the
points in which a conic is intersected by any circle, whose centre is at
a fixed point on the transverse axis, is constant.

17. Shew that the equation to the circle circumscribing the triangle

2a

formed by the three tangents to the parabola r = z drawn at

•^ 1 - cos d

the points whose vectorial angles are a, j3, and y, is



a 8 y . fa+B+y

r=a cosec - cosec ~ cosec ^ sm '



-)



2 2 2 V 2

and hence that it always passes through the focus.

18. If tangents be drawn to the same parabola at points whose
vectorial angles are a, /3, y, and 8, shew that the centres of the circles
circumscribing the four triangles formed by these four Hnes all lie on
the circle whose equation is



a B y 8 r^ a + ^ + y + S-

r= - a cosec ^ cosec ^- cosec ^ cosec - cos ' ^
z A d a



fg a + P+y + 8 -\



19. The circle circumscribing the triangle formed by three tangents
to a parabola is drawn; prove that the tangent to it at the focus
makes with the axis an angle equal to the sum of the angles made
with the axis by the three tangents.



XXXIX.] POLAR EQUATION. EXAMPLES. 321

20. Shew that the equation to the circle, which passes through
the focus and touches the curve - = 1 - ecos 6 at the point ^ = a, is

ril-e cos af = l cos {d -a) - el cos {9 - 2a).

21. A given circle, whose centre is on the axis of a parabola,
passes through the focus S and is cut in four points A, B, C, and D by
any conic, of given latus-rectum, having S as focus and a tangent to-
the parabola for directrix ; prove that the sum of the distances of the
points A, B, G, and D from S is constant.

22. Prove that the locus of the vertices of all parabolas that can be
drawn touching a given circle of radius a and having a fixed point on

the circumference as focus is r=2acos^-, the fixed point being the

pole and the diameter through it the initial line.

23. Two conic sections have the same focus and directrix. Shew
that any tangent from the outer curve to the inner one subtends a
constant angle at the focus.

.24. Two equal ellipses, of eccentricity e, are placed with their
axes at right angles and they have one focus S in common ; if PQ be

g
a common tangent, shew that the angle PSQ is equal to 2 sin-^ —p, •

25. Prove that the two conies — ^l-e^cos^ and - = l-e2Cos(^-a)

will touch one another, if

Zi2 (1 _ e^^) + 1^ (1 - e^^) + ^l^l^e-^e^ cos a = 0.

26. An ellipse and a hyperbola have the same focus S and
intersect in four real points, two on each branch of the hyperbola ; if
rj and r^ be the distances from S of the two points of intersection on
the nearer branch, and r^ and r^^ be those of the two points on the
further branch, and if I and V be the semi-latera-recta of the two
conies, prove that

«+r)(Ui)+(z-.)(i + i)=4.

[Make use of Art. 338.]

a

27. If the normals at three points of the parabola r=a cosec^-,

whose vectorial angles are a, /3, and 7, meet in a point whose vectorial
angle is 5, prove that 25=a + /3 + 7-7r.



L. 21



CHAPTER XV.

GENERAL EQUATION OF THE SECOND DEGREE.
TRACING OF CURVES.



348. Particular cases of Conic Sections. The

general definition of a Conic Section in Art. 196 was that
it is the locus of a point P which moves so that its distance
from a given point S is in a constant ratio to its perpen-
dicular distance FM from a given straight line ZK.

When S does not lie on the straight line ZK, we have
found that the locus is an ellipse, a parabola, or a hyperbola
according as the eccentricity e is <= or > 1.

The Circle is a sub-case of the Ellipse. For the
equation of Art. 139 is the same as the equation (6) of
Art. 247 when h^ = a^, i.e. when e = 0. In this case



The Circle is therefore a



CS=0, and SZ=: - ae = oo
e

Conic Section, whose eccentricity is zero, and whose direc-
trix is at an infinite distance.

Next, let S lie on the straight line ZK, so that S and Z
coincide.

In this case, since

SF=e.FM,
we have

. „^,, FM 1


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