lx+my = 1 cuts (i) are given by the equation

(ax + by ― lx — my)2 = 2λxy ................(ii). The line will touch the conic if the lines (ii) are coincident, the condition for which is

Hence the general equation of a conic touching the four straight lines

is

x = 0, y = 0, lx+my-10, and l'x+m'y-1=0, (ax + by −1)2 = 2λxy ;

the parameters a, b, λ being connected by the two equations

217.

λ = 2 (a−1) (b− m) = 2 (a — l') (b — m').

To find the locus of the centres of conics which touch four given straight lines.

If two of the lines be taken for axes, and the equations of the other two lines be

lx + my-1=0, and l'x+m'y-1=0,

the equation of the conic will be

The centre of the conic is given by the equations

a (ax+by − 1)-λy=0, and b (ax + by − 1) — λxx = 0 ; .. axby, and a (2ax-1)=λy ...................(iii). To obtain the required locus we must eliminate a, b and λ from the equations (i), (ii), and (iii).

From (i) and (iii), we have

:

a (2ax-1)= 2y (a− 1) (b — m) = 2 (a− 1) (by — my); therefore, since ax=by,

Similarly, from (ii) and (iii) we have

a (2l'x+2m'y - 1) = 2l'm'y.

Eliminating a, we obtain the equation of the locus of centres, viz.

2lx+2my-1

Im

21'x+2m'y - 1

I'm

This straight line can easily be shewn to pass through the middle points of the diagonals of the quadrilateral, as it clearly should do, for any one of the diagonals is the limiting form of a very thin ellipse which touches the four lines, and the centre of this ellipse is ultimately the middle point of the diagonal. Hence the middle points of the three diagonals of a quadrilateral are points on the centrelocus of the conics touching the sides of the quadrilateral.

218. All conics touching the axes at the two points where they are cut by the line ax + by - 10 are given by the equation

(ax + by − 1)2 = 2λxy.

The conic will be a parabola if λ be such that the terms of the second degree form a perfect square: the condition for this is

The value λ=0 gives a pair of coincident straight lines, viz. (ax + by − 1)2 = 0.

Hence, for the parabola, λ = 2ab, and the equation of the curve is

(ax + by − 1)2 = 4abxy.

The above equation can be reduced to the form

√ax+ √by = 1.

219. To find the equation of the tangent at any point of the parabola √ax+ √by = 1.

We may rationalize the equation of the curve and then

S. C. S.

16

The result

make use of the formula obtained in Art. 177. may however be obtained in a simpler form as follows. The equation of the line joining two points (x, y) and (x", y') on the curve is

Multiply the corresponding sides of the equations (i)

The equation of the tangent at (x', y') is therefore

To find the equation of the polar of any point with respect to the conic, we must use the rationalized form of the equation of the parabola.

Ex. 1. To find the condition that the line lx+my-1=0 may touch the parabola Jax +√by-1=0.

ах

The equation of the tangent at any point (x', y') is

which is the same as the given equation, if l=

Ex. 2. To find the focus of the parabola whose equation is

√ax+√by=1.

The circle which touches TQ at T and which passes through P will also pass through the focus [see Art. 165 (4), two of the tangents being

coincident]. The two points P, Q are (1, 0)

and 0,
(0, 1).

the focus is on both the circles whose equations are

Therefore

Ex. 3. To find the directrix of the parabola Tax+/by= 1.

The directrix is the locus of the intersection of tangents at right angles; now the line lx+my=1 will be perpendicular to y=0 if,

a b
ι m

m-l cos w=0, and the line will touch if + 1. Therefore the inter

cept on the axis of x made by a tangent perpendicular to that axis is

220.

is on the directrix.

is on the directrix.

x (b+a cos w)+y (a+b cos w) = cos w.

Since the foci of a conic are on its axes, if two conics are confocal they must have the same axes.

will, for different values of X, represent different conics of a confocal system. For the distance of a focus from the

centre is

221. The equation of a system of confocal conics is

If λ is positive the curve is an ellipse.

The principal axes of the curve will increase as increases, and their ratio will tend more and more to equality as A is increased more and more; so that a circle of infinite radius is a limiting form of one of the confocals. If λ be negative, the principal axes will decrease as λ increases, and the ratio I will also decrease as

b2+λ
a2 + λ

increases, so that the ellipse becomes flatter and flatter, until is equal to - b2, when the minor axis vanishes, and the major axis is equal to the distance between the foci. Hence the line-ellipse joining the foci is a limiting form of one of the confocals.

If b+λ is negative, the curve is an hyperbola.