the asymptotes be taken for axes, the equation will be xy+ constant = 0; .. a'b'cos ∞ = 0. Hence a+b-2h cos w = 0. (w). 1. EXAMPLES ON CHAPTER IX. Find the centres of the following curves: (iii) За, За 3x-7xy-6y+3x-9y+50. c' = qu2 9 Find also the equations of the curves referred to parallel axes through their centres. 2. What do the following equations represent? (x+1)(Y-2) (i) xy - 2x + y -2=0. (ii) y3 - 2ay + 4ax = 0. 3. Draw the following curves: (7) 41x2+ 24xу+9y - 130ax - 60ay + 116a2 = 0. a‚a da Shew that if two chords of a conic bisect each other, their point of intersection must be the centre of the curve. 5. Shew that the product of the semi-axes of the conic whose equation is (x-2y+1)+(4x+2y-3)-100, is 1. 6. Show that the product of the semi-axes of the ellipse whose equation is x2 - xy + 2y3 — 2x-6y+ 7 = 0 is and that the equation of its axes is x2-y2 - 2xy + 8y - 8 = 0, 2 7. Find for what value of λ the equation 2x2 + λxy — y3 - 3x+6y-9=0 will represent a pair of straight lines. 191 8. Find the equation of the conic whose asymptotes are the lines 2x + 3y-5=0 and 5x + 3y - 8 = 0, and which passes through the point (1, − 1). 9. Find the equation of the asymptotes of the conic 3x-2xy-5y+7x-9y=0; and find the equation of the conic which has the same asymptotes and which passes through the point (2, 2). 10. Find the asymptotes of the hyperbola 6x27xy-3y-2x-8y-6=0; find also the equation of the conjugate hyperbola. 11. Shew that, if ax2 + 2hxy + by2 = 1, and a'x2 + 2h'xy + b'y2 = 1 represent the same conic, and the axes are rectangular, then (a — b)2 + 4h2 = (a' — b′)2 + 4h'3. 12. Shew that for all positions of the axes so long as they remain rectangular, and the origin is unchanged, the value of g2+ƒ2 in the equation ax2 + 2hxy + by3 + 2gx + 2fy + c = 0 is constant. 13. From any point on a given straight line tangents are drawn to each of two circles: shew that the locus of the point of intersection of the chords of contact is a hyperbola whose asymptotes are perpendicular to the given line and to the line joining the centres of the two circles. 14. A variable circle always passes through a fixed point and cuts a conic in the points P, Q, R, S; shew that 15. If ax2 + 2hxy + by2 = 1, and Ax2 + 2Hxy + By2 = 1 be the equations of two conics, then will ad +bB+ 2hH be unaltered by any change of rectangular axes. CHAPTER X. MISCELLANEOUS PROPOSITIONS, 176. WE have proved [Art. 167] that the curve represented by an equation of the second degree is always a conic. We shall throughout the present chapter assume that the equation of the conic is ax2+2hxy +by2+2gx+2ƒy+c= 0, unless it is otherwise expressed. The left-hand side of this equation will be sometimes denoted by (x, y). 177. To find the equation of the straight line passing through two points on a conic, and to find the equation of the tangent at any point. Let (x', y') and (x”, y′′) be two points on the conic. a (x − x') (x − x') + h {(x − x') (y — y'') + (x − x') (y — y')} +b (y—y') (y — y') = ax2 + 2hxy + by2+2gx+2fy+c ... (i) when simplified is of the first degree, and therefore represents some straight line. If we put = and y=y' in (i) the left side vanishes identically, and the right side vanishes since (x, y) is on the conic. Hence the point (x, y) is on the line (i). So also the point (x", y') is on the line (i). Hence the equation of the straight line through the two points (x', y') and (x”, y′′) is (i), and this reduces to ax (x + x')+hy (x2 + x') + hx (y' + y′′) + by (y′ + y') + 2gx +2ƒy+c=ax'x' + h(x'y′′ + y'x′′)+by' y”.......(ii). To obtain the tangent at (x, y) we put x"=x', and y" = y′ in (ii), and we get 2axx' + 2h (xy' + x'y) + 2byy' + 2gx+2ƒy+c=ax"2 +2hx'y' + by2. Add 2gx+2fy+c to both sides: then, since (x', y') is on the conic, the right side will vanish; and we get for the equation of the tangent ax'x+h (y'x+x'y)+by'y+g(x+x') +ƒ(y+y')+c=0. It should be noticed that the equation of the tangent at (x, y) is obtained from the equation of the curve by writing x'x for x2, y'x+x'y for 2xy, y'y for y2, x+x' for 2x, and y+y for 2y. (w). 178. To find the condition that a given straight line may be a tangent to a conic. Let the equation of the straight line be The equation of the straight lines joining the origin to the points where the line (i) cuts the conic (x, y)=0, are given [Art. 38] by the equation If the line (i) be a tangent it will cut the conic in coincident points, and therefore the lines (ii) must be coincident. The condition for this is S. C. S. 13 (an2 — 2gln + cl2) (bn3 — 2ƒmn + cm3) = (hn2 -fln - gmn + clm)2, or l2 (bc — ƒ2) + m2 (ca — g3) + n2 (ab − h3) + 2mn (gh — fa) + 2nl (hf − gb)+ 2lm (fg — hc) = 0................(iii). The equation (iii) may be written in the form Al2+Bm2 + C'n2 + 2Fmn + 2 Gnl+2Hlm=0...(iv), where the coefficients A, B, C, &c. are the minors of a, b, c, &c. in the determinant 179. To find the equation of the polar of any point with respect to a conic. It may be shewn, exactly as in Article 76, 100, or 118, that the equation of the polar is of the same form as the equation of the tangent, The equation of the polar of (x', y') is therefore ax'x+h (y'x+x'y) +by'y + g (x + x') +ƒ(y+y')+c=0, or x(ax2+hy+g) + y (hx′ + by +ƒ) + gx′ +fy' +c=0. The equation of the polar of the origin is found by putting xy' = 0 in the above; the result is = gx+fy+c=0. 180. If two points P, Q be such that Q is on the polar of P with respect to a conic, then will P be on the polar of Qwith respect to that conic. Let the co-ordinates of P be x', y', and those of Q x", y". The equation of the polar of P is ax'x+ h (y'x + x'y) + by'y + g (x + x') +f(y+y') + c = 0. Since (x", y'') is on the polar of P, we have ax'x"+h (y'x'"+x'y')+by'y'+g(x′+x") +ƒ(y+y′′)+c=0. The symmetry of this result shews that it is also the condition that the polar of Q should pass through P. |