Since (i) and (ii) represent the same curve, and the coefficient of xy is zero in (i), the coefficient of xy must be

Hence a directrix is parallel to one or other of the

axes.

Let a= 0, then since the coefficients of x and y are zero in (i), we have y' = 0 and x' e2p.

=

Also, by comparing the other coefficients in (i) and (ii), we have

From (v) we see that there are two foci on the axis of

x whose distances from the centre are ± √ (1-풍).

From (iv) we see that a directrix is the polar of the corresponding focus.

If a

=

П

2

we can shew in a similar manner that there

are two foci on the axis of y whose distances from the

centre are

1

-). Of the two pairs of foci one is

clearly real and the other imaginary, whatever the values of a and b (supposed real) may be.

The eccentricity of a conic referred to a focus on the

axis of x is from (ii) equal to /(1-2); the eccentricity

referred to a focus on the axis of y will similarly be b

√(1-2). If the curve be an ellipse a and b have the

same sign, and one of these eccentricities is real and the other imaginary. If however the curve be an hyperbola, a and b have different signs and both eccentricities are real.

191. To find the eccentricity of a conic given by the general equation of the second degree.

By changing the axes we can reduce the conic to the

If e be one of the eccentricities of the conic,

..(i).

(ii).

(iii),

..(iv).

=

Eliminating a and ẞ from the equations (ii), (iii) and

(iv), we have

If the curve is an ellipse, ab-h2 is positive, and one value of e is positive and the other negative. The real value of e is the eccentricity of the ellipse with reference to one of the real foci, and the imaginary value is the eccentricity with reference to one of the imaginary foci.

If the curve is an hyperbola both values of e2 are positive, and therefore both eccentricities are real, as we found in Art. 190; we must therefore distinguish between the two eccentricities.

α

The signs of a and ẞ in (i) are different when the curve is an hyperbola; and, if the sign of a be different from

that of y, the real foci will lie on the axis of x. Hence to find the eccentricity with reference to a real focus; obtain the values of a and ẞ from (iii) and (iv), then (ii) will give the eccentricity required, if we take for a that value whose sign is different from the sign of y.

Ex. Find the eccentricity of the conic whose equation is x2-4xy - 2y2+10x+4y=0.

The equation referred to the centre is x2 - 4xy - 2y2-1-0. This will become ax2+By2-1=0, where a+8=-1 and aß=-6. Hence a=2, B=-3. The eccentricity with reference to a real focus is given by 2=−3 (1− e2); therefore e=√

192. To find the foci of a conic.

Let (x, y) be one of the foci of the conic

ax2+2hxy+by2+2gx+2fy+c=0

....................(i). The corresponding directrix of the conic is the polar of (x, y); therefore its equation is

x (ax' +hy+g)+y (hx′ + by' +ƒ) +gx' +ƒy' +c=0. The equation of the conic may therefore be written in the form

(x − x')2 + (y — y')2 — λ {x (ax' + hy' +g)+y (hx' + by' +f) +gx' +fy' + c}=0......(ii). Since (i) and (ii) represent the same curve, the coefficients in (ii) must be equal to the corresponding coefficients in (i) multiplied by some constant. We have therefore

1-λ(ax+hy+g)2

= ka,

- λ (ax' + hy' +g) (hx′ + by' +ƒ) = kh, 1-λ (hx+by+ƒ)2

= kb,

From the first three of the above equations we have

(ax' + hy' +g)2 − (hx′ + by′ +ƒ2)

a-b

(ax' + hy′ +g) (hx′ +by′+ƒ) .......(iii).

h

Multiply the fourth and fifth equations by x, y respectively and add them to the sixth; then, comparing with the second, after rejecting the factor gx' +fy' + c, we get x' (ax′+hy+g)+y' (hx' +by' +ƒ)+gx' +ƒy' +c

(ax' +hy+g) (hx' + by' +ƒ) h

(ax' + hy' + g) (hx′ +by+f).......(iv).

h

The four foci are therefore from (iii) and (iv) the four points of intersection of the two conics

(ax+hy+g)2 — (hx+by+ƒ)2 _ (ax+hy+g) (hx+by+f)

193. The equation of a conic referred to a focus as origin is x2 + y2 = e2 (x cosa + y sin a — p)2.

Either of the lines ±√-1y=0 meets the conic in coincident points.

Hence the tangents from the focus to the conic are the imaginary lines x±y √-1=0, or as one equation

x2 + y2 = 0. ·

Since the equation of the tangents from a focus is independent of the position of the directrix, it follows that if conics have one focus common they have two imaginary tangents common, and that confocal conics have four common tangents.

Now if the origin and axes of co-ordinates be changed in any manner, the equation of the tangents from a focus will be changed from

x2 + y2 = 0 to x2 + y2 + 2gx+2ƒy+c=0.

Hence the equation of the tangents to a conic from a focus satisfies the conditions for a circle.

We may therefore find the foci of a conic in the following manner.

The equation of the tangents from (x', y') to the conic (x, y) = 0 is

(ax2 + 2hxy+by2 + 2gx + 2fy + c) $ (x', y')

=

{ax'x + h(x'y+y'x) + by y + g(x+x') +ƒ (y+y')+c}2. If (x', y') be a focus of the conic, this equation satisfies the conditions for a circle, viz. that the coefficients of x2 and y2 are equal, and that the coefficient of xy is zero.

Hence we have

ap (x', y') − (ax' + hy' +g)2 = b$ (x', y') — (hx' + by +ƒ)2, and hp (x', y') = (ax' + hy' +g) (hx' + by' +ƒ).

The foci are therefore the points given by

(ax +hy+g)2 − (hx + by +ƒ)*

194. To find the equation of the axes of a conic. The axes of a conic bisect the angles between the asymptotes, and the asymptotes are parallel to the lines given by the equation ax2+2hyx + by2=0 [Art. 174]. Hence [Art. 39] the axes are the straight lines through the centre of the conic parallel to the lines given by the equation

may also find the equation of the axes as follows. If a point P be on an axis of the conic, the line joining P to the centre of the conic is perpendicular to the polar of P.

Let x', y be the co-ordinates of P, then the equation. of the polar of P is

x(ax'+hy+g)+y (hx'+by'+f) + gx' +ƒy'+c=0...(i).

S. C. S.

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