hence 4p= 2 √{4 (aλ — g)2 + 4 (Bλ —ƒ)"} Hence (i) is a parabola whose axis is the line ax + By +λ = 0, and whose latus-rectum is 2 √/ {(aλ − g)2 + (Bλ − ƒ)2} _ 2 (af — Bg) 173. We will now find the nature and position of the conics given by the following equations. (1) (1) 7x2-17xy+6y2+23x-2y-20=0. (2) x2-5xy + y2+8x-20y+15=0. (3) 36x2+24xy +29y2 - 72x+126y+81=0. (4) (5x-12y)2 - 2x - 29y-10. The equations for finding the centre are [Art. 168, (i), (ii)] 14x-17y'+23=0) - 17x+12y' −2=0) • These give x=2, y=3. Therefore centre is the point (2, 3). 169] The equation referred to parallel axes through the centre will be [Art. 7x2 - 17xy+6y2+ 23 2 .2-1.3-20=0. or 7x2-17xy+6y2=0. The equation therefore represents two straight lines which intersect in the point (2, 3). They cut the axis of x, where 7x2+23x-20=0, that is where x = 4, and where x= or 5 7་ (2) x2-5xy+y2+8x-20y+15=0. The equations for finding the centre are 2.x' - 5y'+8=0, and -5x+2y' - 20=0; .. x' = — 4, y' = 0. The equation referred to parallel axes through the centre will be x2 - 5xy + y2+ 4 ( − 4)+15=0, x2-5xy+y2=1. or The semi-axes of the conic are the roots of the equation The curve is therefore an hyperbola whose real semi-axis is 14, 7 or or X The direction of the real axis is given [Art. 171, (iv)] by the equation The equations for finding the centre are 36x'+12y'-36=0, and 12x+29y' +63=0; ... x'=2, y'=-3. The equation referred to parallel axes through the centre, will be 36x2+24xy+29y2 - 72+63 ( -3)+81=0, The semi-axes of the conic are the roots of the equation or The equation of the major axis is [Art. 171, (iv)] (5x-12y +λ)2 = 2x (1+5λ) +y (29 − 24λ) +λ2+1. H therefore 5x-12y+1=0 is the equation of the axis of the parabola, and 12x+5y+2=0 is the equation of the tangent at the vertex. Every point on the curve must clearly be on the positive side of the line 12x+5y+2=0, since the left side of equation (i) is always positive. 174. To find the equation of the asymptotes of a conic. We have seen [Art. 146] that the equations of a conic and of the asymptotes only differ by a constant. Let the equation of a conic be ax2 + 2hxy+by2+2gx+2fy + c = 0......(i). Then the equations of the asymptotes will be ax2 + 2hxy + by2+2gx+2ƒy+c+λ = 0.................(ii), provided we give to λ that value which will make (ii) represent a pair of straight lines. The condition that (ii) may represent a pair of straight lines is [Art. 170] ab (c+λ) + 2fgh — af2 — bg2 — (c + λ) h2 = 0 ; ) Hence the equation of the asymptotes of (i) is ax2 + 2hxy + by2 + 2gx + 2fy+c A ab-h2 A == 0. D The equations of two conjugate hyperbolas differ from the equation of their asymptotes by constants which are equal and opposite to one another [Art. 152]; therefore the equation of the hyperbola conjugate to (i) is ax2+2hxy+by+2gx+2fy+c 2A ab-h2 = 0. Cor. The lines represented by the equation ax2 + 2hxy+by2 = 0 are parallel to the asymptotes of the conic, Ex. Find the asymptotes of the conic x2-xy - 2y2+3y-2=0, (w). The asymptotes will be x2-xy - 2y2+3y-2+λ=0, if this equation represents straight lines. Solving as a quadratic in x, we have Hence [Art. 37], the condition for straight lines is 9 (2-1)=9, or λ=1. The asymptotes are therefore x2 - xy - 2y2+3y-1-0. 175. To find the condition that the conic represented by the general equation of the second degree may be a rectangular hyperbola. If the equation of the conic be ax2 + 2hxy+by2+2gx+2fy + c = 0, the equation ax2 + 2hxy + by2 = 0 represents straight lines parallel to the asymptotes. .(i) Hence, if the conic is a rectangular hyperbola, the lines given by (i) must be at right angles. The required condition is therefore [Art. 44] If the axes of co-ordinates be at right angles to one another the condition is The required condition may also be found as follows. If the axes of co-ordinates be changed in any manner whatever, we have a+b-2h cos w a+b' - 2h' cos w' [Art. 52]. sin' w sin2 w' But, if the conic be a rectangular hyperbola and |