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The equation of any line through the centre of the conic is ax+hy+g+λ(hx+by+ƒ) = 0.........................(ii). Since (ii) is perpendicular to (i), we have

(a+λh) (ax'+hy+g)+(h+λb) (hx'+by′+ƒ)=0...(iii). Since (ii) passes through (x', y'), we have

ax' + hy' +g+λ (hx' + by′+ƒ) = 0...................(iv). Eliminate from (iii) and (iv), and we see that (x, y) must be on the conic

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

a-b

which is the equation required.

h

The equation of the axes may also be deduced from Article 192 or 193; for one of the conics on which we have found that the foci lie passes through the centre, and therefore must be the axes.

Ex. 1. Shew that all conics through the four foci of a conic are rectangular hyperbolas.

Ex. 2. Prove that the foci of the conic whose equation is

lie on the curves

ax2+2hxy+by2 = 1,

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are

Ex. 4.

x2-6xу + y2 - 2x - 2y+5=0 are (1, 1) and (-2, − 2).
The co-ordinates of the real foci of 2x2 - 8xy-4y2 — 4y+1=0

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Ex. 5. The focus of the parabola x2+2xy + y2 − 4x+8y − 6=0 is the point (-3).

Ex. 6. Shew that the product of the perpendiculars from the two imaginary foci of an ellipse on any tangent to the curve is equal to the square of the semi-major axis.

Ex. 7. Shew that the foot of the perpendicular from an imaginary focus of an ellipse on the tangent at any point lies on the circle described on the minor axis as diameter.

Ex. 8. If a circle have double contact with an ellipse, shew that the tangent to the circle from any point on the ellipse varies as the distance of that point from the chord of contact.

195. To find the equation of a conic when the axes of co-ordinates are the tangent and normal at any point. The most general form of the equation of a conic is ax2 + 2hxy+by+2gx+2fy + c = 0.

Since the origin is on the curve, the co-ordinates (0, 0) will satisfy the equation, and therefore c = 0.

The line

=

0.

= 0 meets the curve where ax2+2gx y= If y=0 is the tangent at the origin, both the values of x given by the equation ax2+2gx=0 must be zero; therefore g = 0.

Hence the most general form of the equation of a conic, when referred to a tangent and the corresponding normal as axes of x and y respectively, is

ax2+2hxy+by2+2fy = 0.

Ex. 1. All chords of a conic which subtend a right angle at a fixed point O on the conic, cut the normal at O in a fixed point.

Take the tangent and normal at O for axes; then the equation of the conic will be

ax2+2hxy+by2+2fy=0.

Let the equation of PQ, one of the chords, be lx+my -1=0. The equation of the lines OP, OQ will be [Art. 38]

ax2+2hxy+by2+2fy (lx+my)=0

.......(i).

But OP, OQ are at right angles to one another, therefore the sum of the coefficients of x2 and y2 in (i) is zero. Hence we have a+b+2fm=0; which shews that m is constant, and m is the reciprocal of the intercept on the normal.

Ex. 2. If any two chords OP, OQ of a conic make equal angles with the tangent at O, the line PQ will cut that tangent in a fixed point.

196.

The equation of the normal at any point (x, y) of the conic whose equation is ax2 + by2 = 1 is

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This will pass through the point (h, k) if

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Therefore the feet of the normals which pass through a particular point (h, k) are on the conic

xy (a − b) + bhy — akx = 0.................................(i).

The four real or imaginary points of intersection of the conic (i) and the original conic are the points the normals at which pass through the point (h, k).

The conic (i) is clearly a rectangular hyperbola whose asymptotes are parallel to the axes of co-ordinates, that is to the axes of the original conic. It also passes through the centre of that conic, and through the point (h, k) itself.

197. If the normals at the extremities of the two chords le+ my-1=0 and lx+m'y-1=0 meet in the point (h, k), then, for some value of X, the conic

ax2 + by3 − 1 − λ (lx+my − 1) (l'x+m'y − 1) = 0...(i), which, for all values of X, passes through the four extremities of the two chords, will [Art. 196] be the same as

xy (a - b)+bhy-akx=0............(ii).

The coefficients of x and y3, and the constant term are all zero in this last equation, and therefore they must be zero in the preceding.

We have therefore

a-λll = 0, b−λmm' = 0, and 1+λ=0.

Hence, if the normals at the ends of the chords lx + my − 1 = 0 and l'x+m'y−1 =0 meet in a point, we have

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198. By the preceding Article we see that normals to the ellipse whose axes are 2a, 26 at the extremities of the

chords whose equations are

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

will meet in a point, if

a2ll' = b2mm':

=

......

.(i).

If the eccentric angles of these four points be a, ẞ and y, 8, the equations of the chords will be

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We have therefore, by comparing with (i),

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whence

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a+B+y+8= (2n + 1) π.............. (ii).

Also the first equation gives

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and, using the condition (ii), this becomes

sin (a + B) + sin (B+ y) + sin (y + a) = 0...(iii).

Ex. 1. If ABC be a maximum triangle inscribed in an ellipse, the normals at A, B, C will meet in a point.

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condition that the normals meet in a point is [Art. 198 (iii)]

sin 2a+sin( 2a+
(2a + 2) + sin (2a+17) =

which is clearly true.

=0,

Ex. 2. The normals to a central conic at the four points P, Q, R, S meet in a point, and the circle through P, Q, R cuts the conic again in S'; shew that SS' is a diameter of the conic.

SS' will be a diameter of the conic if RS and RS' are parallel to conjugate diameters [Art. 134].

Now if PQ be lx+my-1=0, RS will be

a

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b

m

y+1=0 [Art. 197];

also RS' will be parallel to lx - my=0, since P, Q, R, S' are on a circle;

hence SS is a diameter, for [Art. 182] lx-my=0, and conjugate diameters of ax + by2=1.

a b

7x+y=0 are

m

[The proposition may also be obtained from Art. 198 (ii), and Art. 186, Ex. (1).]

Ex. 3. If the normals to an ellipse at A, B, C, D meet in a point, the axis of a parabola through A, B, C, D is parallel to one or other of the equi-conjugates.

If h, k be the point where the normals meet, A, B, C, D are the four points of intersection of the conics

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All conics through the intersections are included in the equation

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If this be a parabola the terms of the second degree must be a perfect

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axes are therefore [Art. 172] parallel to one or other of the lines+=0.

a

Ex. 4. The perpendicular from any point P on its polar with respect to a conic passes through a fixed point 0; prove (a) that the locus of P is a rectangular hyperbola, (B) that the circle circumscribing the triangle which the polar of P cuts off from the axes always passes through a fixed point O', (y) that a parabola whose focus is O' will touch the axes and all such polars, (8) that the directrix of this parabola is CO, where C is the centre of the conic, and (e) that O and O' are interchangeable.

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