These two chords are equally inclined to the axis by Cor. 4: therefore tan (a+6)=-tan (y+8), or (a +ß)=nπ − } (y +8) ; therefore a+B+y+d=2nr. Ex. 2. A focal chord of a conic varies as the square of the parallel diameter. [See Art. 161.] Ex. 3. If a triangle circumscribe a conic the three lines from the angular points of the triangle to the points of contact of the opposite sides will meet in a point. Let the angular points be A, B, C and the points of contact of the opposite sides of the triangle be A′, B′, C'; also let 1, T T be the semidiameters of the conic parallel to the sides of the triangle. Then BA' : BC'=r1 ÷ rg; CB′ : CA'=r2 : r1; and AC′ : AB'=ï1⁄2 : Tq. BA.CB'. AC=BC", AB', CA', Hence which shews that the three lines meet in a point. Ex. 4. If a conic cut the three sides of a triangle ABC in the points A' and A", B' and B", C' and C" respectively, then will BA, BA". CB". CB" ; AC". AC"=BC", BC". CA' .CA". AB'. AB". (Carnot's Theorem.) [BA',BA" : BC".BC"=r12; r22, and so for the others; 71, 72, 71⁄2 being the semi-diameters of the conic parallel to the sides of the triangle.] 2 Ex. 5. If a conic touch all the sides of a polygon ABCD...... the points of contact of the sides AB, BC...... being P, Q, R, S......; then will AP.BQ.CR.DS...... be equal to PB.QC.RD...... 187. If S be written for shortness instead of the lefthand side of the equation ax2 + 2hxy+by2+2gx+2fy+c=0, and S' be written instead of the left-hand side of the equation = 0, a'x2+2h'xy+b'y2+2g'x + 2ƒ'y + c' then S-AS=0 is the equation of a conic which passes through the points common to the two conics S=0, S'=0. For, the equation S-XS'=0 is of the second degree, and therefore represents some conic. Also if any point be on both the given conics, its co-ordinates will satisfy both the equations S=0 and S'= 0, and therefore also the equation S-λ S'=0. By giving a suitable value to λ, the conic S-XS' = 0 can be made to satisfy any one other condition. If the conic S=0 really be two straight lines whose equations are la+my+n=0 and l'x + m'y + n' = 0, which for shortness we will call u = 0, and v = 0, then S-X uv = 0 will, for all values of λ, be the equation of a conic passing through the points where S = 0 is cut by the lines u = 0 and v= 0. v = If now the line 0 be supposed to move up to and ultimately coincide with the line u=0, the equation S-λu=0 will, for all values of A, represent a conic which cuts the conic S= 0 in two pairs of coincident points, where S=0 is met by the line u=0. That is to say S-λu2=0 is a conic touching S=0 at the two points where S=0 is cut by u = 0. (w). Ex. 1. All conics through the points of intersection of two rectangular hyperbolas are rectangular hyperbolas. If S=0, S'=0 be the equations of two rectangular hyperbolas, all conics through their points of intersection are included in the equation S->S'=0. Now the sum of the coefficients of x2 and y2 in S-XS'=0 will be zero, since that sum is zero in S and also in S', the axes being at right angles. This proves the proposition. [Art. 175.] The following are particular cases of the above. (i) If two rectangular hyperbolas intersect in four points, the line joining any two of the points is perpendicular to the line joining the other two. (For the pair of lines is a conic through the points of intersection.) (ii) If a rectangular hyperbola pass through the angular points of a triangle it will also pass through the orthocentre. (For, if A, B, C be the angular points, and the perpendicular from A on BC cut the conic in D; then the pair of lines AD, BC is a rectangular hyperbola, since these lines are at right angles; therefore the pair BD, AC is also a rectangular hyperbola, that is to say the lines are at right angles.) Ex. 2. If two conics have their axes parallel a circle will pass through their points of intersection. Take axes parallel to the axes of the conics, their equations will then be and ax2+by2+2gx+2fy+c=0, a'x2+b'y2+2g'x+2f'y+c'=0. The conic ax2 + by2+2gx+2fy+c+λ (a'x2+b'y2+2g'x+2f'y+c')=0 will go through their intersections. But this will be a circle, if we choose λ so that a +λa'=b+λb', and this is clearly always possible. Ex. 3. If TP, TQ and T'P', T'Q' be tangents to an ellipse, a conic will pass through the six points T, P, Q, T', P', Q. Let the conic be ax2+by2=1, and let T be (x', y') and T' be (x", y'). The equations of PQ and P'Q' will be axx'+byy' -1=0 and axx" + byy" -1=0. The conic λ (ax2 + by2 − 1) − (axx' + byy' —1) (axx" + byy′′ − 1)=0 will always pass through the four points P, Q, P', Q'. It will also pass through T if λ be such that or if λ (ax'2 + by'2 – 1) − (ax22 + byˆ2 - 1) (ax'x" + by'y′′ − 1) =0, The symmetry of this result shews that the conic will likewise pass through T'. Ex. 4. If two chords of a conic be drawn through two points on a diameter equidistant from the centre, any conic through the extremities of those chords will be cut by that diameter in points equiḍistant from the centre. Take the diameter and its conjugate for axes, then the equation of the conic will be ax2+by2=1. Let the equations of the chords be y-m(x-c)=0 and y-m' (x+c)=0. Then the equation of any conic through their extremities is given by ax2 + by2 − 1 − λ {y − m (x − c)} {y − m' (x + c)} =0. The axis of x cuts this in points given by ax2 - 1-\mm' (x2 — c2)=0, and these two values of x are clearly equal and opposite whatever A, m and m' may be. As a particular case, if PSQ and P'S'Q' be two focal chords of a conic, the lines PP' and QQ' cut the axis in points equidistant from the centre. 188. To find the equation of the pair of tangents drawn from any point to a conic. Let the equation of the conic be ax2 + 2hxy + by3 + 2gx+2fy + c = 0................(i). If (x, y) be the point from which the tangents are drawn, the equation of the chord of contact will be axx' + h (xy' + yx') +byy' +9 (x + x') +ƒ (y + y) +c=0. The equation ax2+2hxy + by2+2gx + 2fy + c =λ {axx' +h (xy' + yx') + byy' + g(x + x') +ƒ (y+y') +c}2 ......(ii) represents a conic touching the original conic at the two points where it is met by the chord of contact. The two tangents are a conic which touches at these two points and which also passes through the point (x, y) itself. The equation (ii) will therefore be the equation required if › be so chosen that (x', y') is on (ii); that is, if ax2 + 2hx'y' + by'2 + 2gx' + 2ƒy + c Therefore 1=λ {ax^2 + 2hx'y' + by'2 + 2gx' + 2fy' + c} = λp (x', y'). λφ Substituting this value of λ in (ii) we have = (ax2 + 2hxy + by2 + 2gx + 2fy + c) $ (x', y') = {axx' + h (xy' + yx') + byy' + g (x + x) +ƒ (y+y') +c}3, which is the required equation. The above equation may be found in the following manner. (w). Let TQ, TQ' be the two tangents from (x', y'), let P (x, y) be any point on TQ, and let TN, PM be the perpendiculars from T and P on the chord of contact QQ'. and [Art. 31] PM2 TN2 .(i). {axx' + h (xy' +yx') + byy' + g (x+x') +ƒ (y+y')+c}2. therefore from (i) we have ; • (x, y) + (x', y') = {axx' + h (xy' + yx')+byy' + g (x+x') + f (y+y')+c}2. 189. To find the equation of the director-circle of a conic. The equation of the tangents drawn from (x, y) to the conic given by the general equation is = (ax2 + 2hxy + by2 + 2gx + 2fy + c) $ (x', y) {axx' + h(xy + yx') + byy + g(x+x) + ƒ (y + y) + c}*. The two tangents will be at right angles to one another if the sum of the coefficients of x and y2 in the above equation is zero. This requires that (a+b)(ax2 + 2hx'y' + by'2 + 2gx' + 2fy' + c) − (ax' + hy' + g)3 — (hx' + by' +ƒ)* = 0. The point (x, y) is therefore on the circle whose equation is (ab — h2) (x2 + y2) + 2x (gb −fh) + 2y (fa − hg) + c (a + b) or Cx2 + Cy2 — 2Gx − 2Fy+A+B=0 ....(i), ...... where A, B, C, F, G, H mean the same as in Art. 178, If h2 - ab=0, the equation reduces to or 2x (bg − ƒh) + 2y (fa − hg) + c (a + b) − ƒa − g2 = 0, 2Gx+2Fy-A-B = 0..................................(ii). The conic in this case is a parabola, and (ii) is the equation of its directrix. Ex. 1. Trace the curve 11x2 +24xy + 4y2 - 2x+16y+11=0, and shew that the equation of the director-circle is x2+ y2+2x-2y=1. Ex. 2. Shew that the equation of the directrix of the parabola x2+2xy + y2 - 4x+8y-6=0 is 3x-3y+8=0. 190. To shew that a central conic has four and only four foci, two of which are real and two imaginary. Let the equation of the conic be ax2 + by2 − 1 = 0.......... .(i). Let (x, y) be a focus, and let x cos a+ y sin a—p=0 be the equation of the corresponding directrix; then if e be the eccentricity of the conic, the equation will be (x − x')2 + (y — y')2 — e2 (x cos a + y sin x-p)2 = 0...(ii). |