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S. L. (Sidney Luxton) Loney.

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and the point in which it cuts the polar of 0.

Ex. Through any point is draion a straight line to cut a conic
in P and P' and on it is taken a point R such that OR is (1) the
arithmetic mean, and (2) the geometric mean, betioeen OP and OP'.
Find in each case the locus of R.

Using the same notation as in the last article, we have

0P+0P'=-2 - g COB d+f Bind

and OP.OP'=



a cos2 + 2h cos ^ sin ^ + & sin^ d *
c



a cos^ 6 + 2h cos 6 sin ^ + & sin^ '



374 COORDINATE GEOMETRY.

(1) If R be the point (p, 6) we have

-i(np,oP'\- g COB d+f sine

p-^[^ujr-tLjjr ) acos'^d + 2hcoseBmd + bs'm^d'

i.e. ap cos2 d + 27ip cos 6 Bin 6 + hp sin^ d + gQOBd-\-f sin ^ = 0,
i.e., in Cartesian coordinates,

ax^ + 2 7ia;i/ + by^+gx +fy = 0.
The locus is therefore a conic passing through and the inter-
section of the conic and the polar of 0, i.e. through the points T
and T', and having its asymptotes parallel to those of the given
conic.

(2) If R be the point {p, 6), we have in this case

/•

„2 — np OP'—

^ ^^.^ acos2^ + 27icos^sin0 + &sin2 6''

i.e. ap^ cos^ d + 2hp^ cos ^ sin ^ + hp- sin^ ^ = c,

i.e. aa:2 + 2/ia;^ + 6?/2 = c.

The locus is therefore a conic, having its centre at and passing
through T and T', and having its asymptotes parallel to those of the
given conic.

402. To find the locus of the middle points of 'parallel cliords of a
conic. [Cf. Art. 376.]

The lengths of the segments of the chord drawn through the point
{x\ y') at an angle 6 to the axis of x is given by equation (2) of Art.
397.

If {x', y') be the middle point of the chord the roots of this
equation are equal in magnitude but opposite in sign, so that their
algebraic sum is zero.

The coefi&cient of r in this equation is therefore zero, so that
{ax' + hy' + g) cos + [hx' + by' +/) sin ^ = 0.

The locus of the middle point of chords inclined at an angle 6 to
the axis of x is therefore the straight line

(ax + hy + g) + {hx + by +f) tan ^ = 0.
Hence the locus of the middle points of chords parallel to the line
y = mx is

{ax + hy +g) + {hx + by+f) m = 0,

i.e. x{a + hm) + {h + bm) y + g +fm = 0.

This is parallel to the line y = m'x if

, a + hm
m = -J- — ^ ,
h + bm

i.e. if a + h{m + m') + bmm' = 0.

This is therefore the condition that y = mx and y — m'x should be
parallel to conjugate diameters.



EXAMPLES. 375

403. Equation to the pair of tangents drawn from a given point
{x', y') to a given conic, [Cf. Art. 389.]

If a straight line be drawn through [x', y'), the point P, to meet
the conic in Q and Q\ the lengths of PQ and PQ' are given by the
equation

r2 (a cos^ ^ + 2/4 cos ^ sin ^ + 6 sin^ 6)

+ 2r l{ax' + hy' + g) cos 6 + {hx' + 5?/' +/) sin ^] + (x', ?/') = 0.

The roots of this equation are equal, i.e. the corresponding Hnes
touch the conic, if

{a cos2 ^ + 2/i cos ^ sin ^ + & sin^ ^) x <^ {x', y')

= [{ax' + hy' + £f) cos 6 + {hx' + by'+f) sin ^p,

I.e. if (a + 27i tan ^ + & tan^ ^) x (a;', y')

= [{ax' + hy' + g) + {hx' + by'+f) tan ef...{l).

The roots of this equation give the corresponding directions of the
tangents through P.

Also the equation to the line through P inclined at an angle 6 to
the axis of x is

^^ = tan0 (2).

x-x

If we substitute for tan 6 in (1) from (2) we shall get the equation
to the pair of tangents from P.
On substitution we have
{aix-x')^ + 2h{x-x'){y-y') + b{y-y'y~}<t>{x',y')

= l{ax' + hy' + g){x- x') + {hx' + by' +/) {y - y')f.
This equation reduces to the form of Art. 389.

EXAMPLES. XLIV.

1. Two tangents are drawn to an ellipse from a point P; if the
points in which these tangents meet the axes of the ellipse be
concyclic, prove that the locus of P is a rectangular hj^erbola.

2. A pair of tangents to the conic Ax^ + By^=l intercept a
constant distance 2 A; on the axis of x ; prove that the locus of their
point of intersection is the curve

By^{Ax^ + By^-l) = Ak^{By^-lf.

3. Pairs of tangents are drawn to the conic ax^ + ^y^=l so as to
be always parallel to conjugate diameters of the conic

ax^ + hxy + by^ = l;

shew that the locus of their point of intersection is the conic

ax" + 2hxy + by^= - + 7; .
a p



376 COORDINATE GEOMETRY. [Exs.

4. Prove that the director circles of all conies which touch two
given straight lines at given points have a common radical axis.

5. A parabola circumscribes a right-angled triangle. Taking its
sides as the axes of coordinates, prove that the locus of the foot of the
perpendicular from the right angle upon the directrix is the curve
whose equation is

2xy {x^ + 2/2) [hy + kx) + hhj^ + A;V= 0,

and that the axis is one of the family of straight lines

m^h - k
y = mx-— K ,

where m is an arbitrary parameter and 2h and 2k are the sides of the
triangle.

Find the foci of the curves

6. 300a;2 + d20xy + 144 </2 - 1220a; - 768?/ + 199 =0.

7. l&x^-2Axy + 9y^ + 28x+14y + 21 = 0.

8. 144.C2 _ i20xy + 25y^ + Qlx - A2y + 13 = 0.

9. x^- &xy + y^- 10a; - lOy -19 = and also its directrices.

10. Prove that the foci of the conic

ax'^ + 2hxy + hy^ = 1
are given by the equations

a-h ~ h ~~ a^-b^'

11. Prove that the locus of the foci of all conies which touch the
four lines x= ^a and y=±b is the hyperbola x^-y^=a^- h^.

12. Given the centre of a conic and two tangents ; prove that the
locus of the foci is a hyperbola.

[Take the two tangents as axes, their inclination being w; let

(^ij Vi) ^^^ {^2> 2/2) ^^ *^^ ^^^^> ^^^ (^' ^) *^® given centre. Then
x-^ + X2=2h and 2/1 + 1/2 = 2^; also, by Art. 270 {j3), we have

y-^y^ sin^ w = x-^x^ sin^ w = (semi-minor axis)^.

From these equations, eliminating x.2 and 2/2, we have

^1^ - y^=2hx^ - 2ky^ ,]

13. A given ellipse, of semi-axes a and h, slides between two
perpendicular lines ; prove that the locus of its focus is the curve

(a;2 + ^2) ^^^2^2 _j. 54) ^ 4aPx^y\

14. Conies are drawn touching both the axes, supposed oblique, at
the same given distance a from the origin. Prove that the foci lie
either on the straight line x = y, or on the circle

x^ + y'^ + 2xycos o}=a{x + y).

15. Find the locus of the foci of conies which have a common point
and a common director circle.



XLIV.] TANGENT AND NORMAL AS AXES. 377

16. Find the locus of the focus of a rectangular hyperbola a
diameter of which is given in magnitude and position.

17. Through a fixed point chords POP' and QOQ' are drawn at
right angles to one another to meet a given conic in P, P' , Q, and Q'.

Prove that ^^, + QO^' ^' ^on^i^ni.

18. A point is taken on the major axis of an ellipse whose abscissa
is ae ~- fj2 — e^ ; prove that the sum of the squares of the reciprocals
of the segments of any chord through it is constant.

19. Through a fixed point is drawn a line OPP' to meet a conic
in P and P' ; prove that the locus of a point Q on OPP', such that

7-rs = Tvvio + y^T^ is another conic whose centre is O.
OQ^ OP^ OP^

20. Prove Carnot's theorem, viz. : If a conic section cut the side
BC of a triangle ABC in the points A' and A'\ and, similarly, the
side CA in B' and B", and AB in C and C'% then

BA' . BA" . GB' . GB" .AG'. AG"=GA' . GA" . AB' . AB" . BG' . BG".

[Use Art. 398.]

21. Obtain the equations giving the foci of the general conic by
making use of the fact that, if >Sf be a focus and PSP' any chord of

the conic passing through it, then -^^^ + — ^, is the same for all direc-

oP hP

tions of the chord.

22- Obtain the equations for the foci also from the fact that the
product of the perpendiculars drawn from them upon any tangent is
the same for all tangents.

404. To find the equation to a conic, the axes of co-
ordinates being a tangent and normal to the conic.

Since the origin is on the curve, the equation to the
curve must be satisfied by the coordinates (0, 0) so that the
equation has no constant term and therefore is of the form
ax^ + 2hxy + h'lf + 2gx + yy — 0.

If this curve touch the axis of x at the orio-in, then,
when 2/ = 0, we must have a perfect square and therefore

The required equation is therefore

ax^ + Ihxy + hy"^ + 2/^/ = (1).

Ex. O is any point on a conic and PQ a chord ; prove that
(1) if PQ subtend a right angle at O, it passes through a fixed
point on the normal at O, and



378



COORDINATE GEOMETRY.



(2) if OP and OQ be equally inclined to the normal at 0, then
PQ passes through a fixed point on the tangent at 0.

Take the tangent and normal at as axes, so that the equation to
the conic is (1).

Let the equation to PQ be y = mx + c (2).

Then, by Art. 122, the equation to the lines OP and OQ is

c {ax'^ + 2hxy + by^) + 2fy {y -mx) = (3).

(1) If the lines OP and OQ be at right angles then (Art. 66), we
have ac + bc + 2/= 0,



I.e.



a + b



= a constant for all positions of PQ.

But c is the intercept of PQ on the axis of y, i.e. on the normal
at 0.

The straight line PQ therefore passes through a fixed point on the

— 2/

normal at which is distant v from O.

a + b

This point is often called the Fregier Point.

(2) If again OP and OQ be equally inclined to the axis of y then,
in equation (3), the coefficient of xy must be zero, and hence

2;ic-2/m = 0,

^ I-
m h

- c



I.e.



= ^ = constant.



But — is the intercept on the axis of x of the line PQ.



m



Hence, in this case, PQ passes through a fixed point on the tangent
at 0.

405. General equation to conies 2^<^ssing through four
given points.

Let A, B, C, and D be the four points, and let BA and
CD meet in 0. Take OAB
and ODC as the axes, and
let OA = \,0B = V, OD - ix,
B.ndOC^fx\

Let any conic passing
through the four points be

ax^i-2h'xi/ + by^

+ 2gx+2/y+c=0...{l).

If we put y = in this
equation the roots of the
resulting equation must be A and A,'.




comes THROUGH FOUR POINTS. 379

Hence 2g = — a (X + X') and c — aX\\

Similarly h — —, , and %f— — c - — y- .

On substituting in (1) we have
fijULof + 2hxi/ + XX'y^ — fxfjL (X + X') x

-XX' {fx + ix)y + XX'fifx' = (1),

where h = h' ^-^ .

G

This is the required equation, h being a constant as yet
undetermined and depending on which of the conies through
A, B, C, and D we are considering.

406. Aliter. We have proved in Art. 383 that the
equation hLN=MR^ ^ being any constant, represents any
conic circumscribing the quadrilateral formed by the four
straight lines L = 0, 31=0, iV= 0, and E = taken in this
order.

With the notation of the previous article the equations
to the four lines AB, BC, CD, and DA are

2/=0, l+y-,-l=0, x = 0,
X fx

and %^_1 = 0.

A fJi

The equation to any conic circumscribing the quadri-
lateral A BCD is therefore

^ - K^^O(^^o w,

i.e.

/xfxx"^ + xy (X/jl' + X'/ji - hXX'ixii) + XX' y"^

— n-ix! (X + X') x~ XX' (fji + fx) y + XX'fjbfji = 0.

On putting Xfx + X'jx - kXX'ixfji' equal to another constant
2h we have the equation (1) of the previous article.



380 COORDINATE GEOMETRY.

407. Only one conic can he drawn through any Jive
points.

For the general equation to a conic through four points
is (1) of Art. 405.

If we wish it to pass through a fifth point, we substitute
the coordinates of this fifth point in this equation, and thus
obtain the corresponding value of h. Except when three of
the five points lie on a straight line a value of h will always
be found, and only one.

Ex. Find the equation to the conic section which passes through
the five points A, B, C, D, and E, whose coordinates are (1, 2), (3, —4),
(-1, 3), (-2, -S),and{5, 6).

The equations to AB, BC, CD, and DA are easily found to be

y + 3x-5 = 0, 4:y + 7x-5 = 0, 6x-y + 9 = 0, and 5x-Sy + l = 0.

The equation to any conic through the four points A, B, C, and D
is therefore

{y + Sx-5){6x-y + 9) = \{4:y + 7x-5){5x-3y + l) (1).

If this conic pass through the point E, the equation (1) must be
satisfied by the values x = 5 and y = &.

We thus have X = V- ^^^i on substitution in (1), the required
equation is

223a;2 - 38xy - 123?/ - 171a; + 8Sy + 350 = 0,

which represents a hyperbola.

4:03. To find the general equation to a conic section
which touches four given straight lines, i.e. which is inscribed
in a given quadrilateral.




THE CONIC LM = R~. 381

Let the four straight lines form the sides of the quadri-
lateral ABCD. Let BA and CD meet in 0, and take OAB
and ODC as the axes of x and y, and let the equations to
the other two sides BG and DA be

l-^x + 7n^y -1=0, and l^x + m^y -1 = 0.

Let the equation to the straight line joining the points
of contact of any conic touching the axes at P and Q be

ax + hy —1 = 0.

By Art. 385, II, the equation to the conic is then

2Xxy = (ax + hy — Vf (1).

The condition that the straight line BG should touch
this conic is, as in Art. 374, found to be

A. = 2(a-Zi)(6-mi) (2).

Similarly, it will be touched by AD if

X = 2{a-l){h-m^) (3).

The required conic has therefore (1) as its equation, the
values of a and h being given in terms of the quantity A. by
means of (2) and (3).

Also X is any quantity we may choose. Hence we have
the system of conies touching the four given lines.

If we solve (2) and (3), we obtain

2& - (mi + W2 ) _ _ 2a-(Zi + y _ ^ / -^ 2\

m^ -7112 ~~ h~h ~ ^ (^1 ~ ^2) ("*i ~ ^^2) *

409. The conic LM=R", where L = 0, M=0, and
R — are the equations of straight lines.

The equation LM=0 represents a conic, viz. two straight
lines.

Hence, by Art. 385, II, the equation

LM=R' (1),

represents a conic touching the straight lines Z = 0, and
M = 0, where R = meets them.



382 COORDINATE GEOMETRY.

Thus L — and M=0 are a pair of tangents and jS =
the corresponding chord of contact.

Every point which satisfies the equations M— jx^L and
B — [xL clearly lies on (1).

Hence the point of intersection of the straight lines
M— fjL^L and E — fxL lies on the conic (1) for all values of
fx. This point may be called the point "ju.."

410. To find the equation to the straight line joining
two jyoints "/x" and "/x"' and the equation to the tangent at
the ^point "/x."

Consider the equation

aL + hM-\-R^O (1).

Since it is of the first degree and contains two constants
a and 6, at our disposal, it can be made to represent any
straight line.

If it pass through the point " /x " it must be satisfied by
the substitutions M = fxL and B — fxL.

Hence a + 6/x^ + /x = (2).

Similarly, if it pass through the point " /x' " we have

a + bfx'' + fM' = (3).

Solving (2) and (3), we have

a , — 1

/X/X fX + fX

On substitution in (1), the equation to the joining line is

Lfxfx' + M-(ix + [x')JR = 0.

By putting fx' = fx we have, as the equation to the
tangent at the point "/x,"

Lfx^ + M- 2fxE = 0,



EXAMPLES. 383



EXAMPLES. XLV.

1. Prove that the locus of the foot of the perpendicular let fall
from the origin upon tangents to the conic ax^ + 2hxy + by' = 2x is the
curve {h^ - ah) {x^ + y'^f + 2 {x^ + ^2) {p^ _ /j^) +^f = Q,

2. In the conic ax'^ + 2hxij + by^ = 2y, prove that the rectangle

contained by the focal distances of the origin is — — - „ .

ao — h^

3. Tangents are drawn to the conic ax^ + 2hxy + by'^ = 2x from
two points on the axis of x equidistant from the origin; prove that
their four points of intersection lie on the conic hy^ + hxy=x.

If the tangents be drawn from two points on the axis of y equi-
distant from the origin, prove that the points of intersection are on a
straight Hne.

4. A system of conies is drawn to pass through four fixed points;
prove that

(1) the polars of a given point all pass through a fixed point,

and (2) the locus of the pole of a given line is a conic section.

5. Find the equation to the conic passing through the origin and
the points (1, 1), ( -1, 1), (2, 0), and (3, -2). Determine its species.

6. Prove that the locus of the centre of all conies circumscribing
the quadrilateral formed by the straight lines ^ = 0, x = 0, x-{-y = l,
and 2/ - a;= 2 is the conic 2x^ - 2y^ + Axy + 5?/ - 2 = 0.

7. Prove that the locus of the centres of all conies, which pass
through the centres of the inscribed and escribed circles of a triangle,
is the circumscribing circle of the triangle.

8. Prove that the locus of the extremities of the principal axes of
all conies, which can be described through the four points ( =t a, 0) and
(0, ± h), is the curve



($.-t)(''^^y')=^'-y-



9. A, B, C, and D are four fixed points and AB and CD meet in
O ; any straight line passing through O meets AD and BC in R and
jR' respectively, and any conic passing through the four given points
in S and S' ; prove that

J^ J^_ J^ , 1

OR "^ OR' ~ OS ''" OS' '

10. Prove that, in general, two parabolas can be drawn through
four points, and that either two, or none, can be drawn.

[For a parabola we have 7i= ± ^JXK'fxfx'.]



384 COORDINATE GEOMETRY. [ExS. XLV.

11. Prove that the locus of the centres of the conies ch'cumserib-
ing a quadrilateral ABGD (Fig. Art. 405) is a conic passing through
the vertices 0, L, and M of the quadrilateral and through the middle
points of AB, AC, AD, BG, BD, and CD.

Prove also that its asymptotes are parallel to the axes of the
parabolas through the four points.

[The required locus is obtained by eliminating h from the equa-
tions 2fjufjL'x + 2hy - fifx,' {\ + \')=0, and 2hx + 2X\'y -XV {fjt. + fM') = 0.]

12. By taking the case when XX'= -/">«•' and when AB and CD
are perpendicular (in which case ABC is a triangle having D as its
orthocentre and AL, BM, and CO are the perpendiculars on its
sides), prove that all conies passing through the vertices of a triangle
and its orthocentre are rectangular hyperbolas.

From Ex, 11 prove also that the locus of its centre is the nine
point circle of the triangle.

13. Prove that the triangle OML (Fig. Art. 405) is such that each
angular point is the pole of the opposite side with respect to any
conic passing through the angular points A, B, C, and D of the
quadrilateral.

[Such a triangle is called a Self Conjugate Triangle.]

14. Prove that only one rectangular hjrperbola can be drawn
through four given points. Prove also that the nine point circles of
the four triangles that can be formed by four given points meet in a
point, viz. , the centre of the rectangular hyperbola passing through
the four points.

15. By using the result of Art. 374, prove that in general, two
conies can be drawn through four points to touch a given straight
line.

A system of conies is inscribed in the same quadrilateral ; prove
that

16. the locus of the pole of a given straight line with respect to
this system is a straight line.

17. the locus of their centres is a straight line passing through the
middle points of the diagonals of the quadrilateral.

18. Prove that the triangle formed by the three diagonals OL,
AC, and BD (Fig. Art. 408) is such that each of its angular points is
the pole of the opposite side with respect to any conic inscribed in the
quadrilateral.

19. Prove that only one parabola can be drawn to touch any four

given lines.

Hence prove that, if the four triangles that can be made by four
lines be drawn, the orthocentres of these four straight lines lie on a
straight line, and their circumcircles meet in a point.



CHAPTER XYII.

MISCELLANEOUS PROPOSITIONS.

On the four normals that can be drawn from any point in
the plane of a central conic to the conic.

411, Let the equation to the conic be

Ax' + By^^l (1).

[If A and JB be both positive, it is an ellipse ; if one be
positive and the other negative, it is a hyperbola.]

The equation to the normal at any point (x, y') of the
curve is

X — X y — y'

If this normal pass through the given point (/«,, k), we
have

h — x k — y'

i.e. (A-£)x'y' + £hy'-Akx'=:0 (2).

This is an equation to determine the point (cc', y') such
that the normal at it goes through the point (h, k). It
shews that the point (x, y') lies on the rectangular hyper-
bola

{A-JB)xy + £hy-Akx = (3).

The point (x', y') is therefore both on the curve (3) and
on the curve (1). Also these two conies intersect in four
points, real or imaginary. There are therefore four points,

L. 25



386 COORDINATE GEOMETRY.

in general, lying on (1), such that the normals at them pass
through the given point (A, k).

Also the hyperbola (3) passes through the origin and
the point (A, k) and its asymptotes are parallel to the axes.

Hence From a given point four normals can in general
he drawn to a given central conic, and their feet all lie on a
certain rectangular hyperbola, which jjasses through the
given point and the centre of the conic, and has its asym^ptotes
parallel to the axes of the given conic.

412. To find the conditions that the normals at the
points where two given straight lines meet a central conic
may m,eet in a p)oint.

Let the conic be

A^-\-By''^\ (1),

and let the normals to it at the points where it is met by
the straight lines

l^ + m,^ — \ (2),

and l^x + m^y = 1 (3)

meet in the point (h, k).

By Art. 384, the equation to any conic passing through
the intersection of (1) with (2) and (3) is

Aa^ + By^ - 1 + X {l-^x + m^y — 1) (^2^ + m.^y - 1) = 0...(4).

Since these intersections are the feet of the four
normals drawn from (A, k), then, by the last article, the
conic

(A- £) xy + Bhy - Akx = (5)

passes through the same four points.

For some value of X it therefore follows that (4) and (5)
are the same.

Comparing these equations, we have, since the co-
efficients of a^ and y^ and the constant term in (5) are all
zero,

A + Xl^l^ = 0, B + Xm^m^ ~ 0, and - 1 +X = 0.

Therefore X =^ 1, and hence

lih — ~^i and m^m^ — ~B (6).



CONCURRENT NORMALS. 387

The relations (6) are the required conditions.

Also, comparing the remaining coefficients in (4) and (5),
we have

A-B -Ak " Bh '

7 A—B m-, + mo ,^.

so that h = ^— y-^ — j-^ (7),

B L^m^ + t^^frix

and ],-_.^lz1 h±^ (8).

Cor. 1. If the given conic be an ellipse, we have
-4 = -5 and B = j^.

The relations (6) then give

cH^^ = IP'm^^ = — 1 (9),

and the coordinates of the point of concurrence are
j^^ a^-h'' m^ + m, ^^ 1 - 5^??^l^

and k = 7^ — -^ j - m-^ (a^ - O'') . —j- — 75 — ^.

Cor. 2. If the equations to the straight lines be given
in the form y = m^c + c and 3/ = m'x + c', we have

m — ^, c= — , m = , and c — — .

mi 7/^1 mg ma

The relations (9) then give



77iy/i' = — „ and cc' = — 6^.



413. If f/ie normals at four points P, Q, R, and S of an ellipse
meet in a point, the sum of their eccentric angles is equal to an odd
multiple of two Hght angles.

25—2



388 COORDINATE GEOMETRY.

If a, /3, 7, and 5 be the eccentric angles of the four points, the
equations to JPQ and BS are

cos



.y=.-a..-cot^-+— — — .
sm-^

& cos '-^

and 2/ = - a^ • - cot ^- + i- . [Art. 259.]

a 2 . 7+0

sm^-

Since the normals at these points meet in a point, we have, by
Art. 412, Cor. 2,

_=mm'=-5 cot —^ cot ^hr -
a^ a^ 2 2



. a + iS 7 + 5 ,

*. tan — ^ = cot = tan

a a



(tt_ _ 7+l5\
V2 "27



a+/3 TT 7 + 5

,. __=n7r+ - -— ,

*.e. a + /3 + 7 + 5=(2w + l)7r.

414. XSx. 1. I/' i/ie normals at the points A, B, C, and D of an

ellipse meet in a point O, prove that SA . SB . SO. SD = \^ . SO^y where
S is one of the foci and X is a constant.

Let the equation to the ellipse be

^i+f:=i (1),

a^ 0'^
and let be the point {h, k).

As in Art. 411, the feet of the normals drawn from lie on the
hyperbola



\a^ b^J



hy kx -



i.e. a^e^xy = aViy — b^kx (2).

The coordinates of the points A, B, G, and D are therefore found
by solving (1) and (2),

From (2) we have y = -o-n 5-\ •

^ ' ^ a^{h-e^x)

Substituting in (1) and simplifying, we obtain

a;4a2g4 _ 2ha^e^x^ + x^ {a^h^ + 62A;2 _ a*e^) + 2;ie2^4^ _ ^4/^2 _ o. . . (3).



CONCURRENT NORMALS. EXAMPLES. 389

If a?!, x^, x^, and x^ be the roots of this equation, we have (Art. 2),



_ 2ha^ _ a^h^



If S be the point ( - ae, 0) we have, by Art. 251,

SA=.a + ex-^.
:. SA.SB.SG.SD = {a + eXj){a+ex.2){a + ex^){a + ex^) '
= a^ + a^Sa?! + a^e^^x^x^ + ae^'Lx-^x^.^ + e'^x^x^^^^

= -^ {(^ + ae)2 + /c2}-, on substitution and simplification,

= k.SOK

Aliter. If p stand for one of the quantities *S^^, SB, SG, or SD
we have p=a + ex,

i.e. x=-{p-a).

Substituting this value in (3) we obtain an equation in the fourth
degree, and easily have

PiPzPsPi^ -2[{h + ^^)^ + ^% ^3 before.

Ex. 2. If the normals at four points P, Q, B, and S of a central
conic meet in a point, and if PQ pass through a fixed point, find the
lotus of the middle point of BS.

Let the equation to PQ be

y = m-^x + Cj^ (1),

and that to BS y = m^-\-C2 ■ (2).

If the equation to the given conic be Ax'^ + By^=l, we then have


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