374. To find the equation of the pair of tangents at the points where S is cut by any line λx + y + vz. The equation of any conic having double contact with S, at the points where it meets. this line, being kS+ (λx + μy + vz)2 = 0, it is required to determine k so that this shall represent two right lines. Now it will be easily verified that in this case not only A' vanishes but also. And if we denote by the quantity Aλ2 + Bμ2 + Cv2 + 2 Fμv + 2 Gvλ + 2Hλμ, the equation to determine k has two roots = 0, the third root being given by the equation kA+2=0. The equation of the pair of tangents is therefore ES=▲ (λx + μy + vz). It is plain that when xx+ μy + vz touches S, the pair of tangents coincides with Ax+y+vz itself; and the condition that this should be the case is plainly Σ=0; as is otherwise proved (Art. 151). Under the problem of this Article is included that of finding the equation of the asymptotes of a conic given by the general trilinear equation. 375. We now examine the geometrical meaning, in general, of the equation =0. Let us choose for triangle of reference any self-conjugate triangle with respect to S, which must then reduce to the form ax + by2+ cz2 (Art. 258). We have therefore ƒ=0,g=0, h=0. The value then of ℗ (Art. 370) reduces to bca' + cab' + abc', and will evidently vanish if we have also a' = 0, b' = 0, c' = 0, that is to say, if S', referred to the same triangle, be of the form f'yz+g'zx+h'xy. Hence vanishes whenever any triangle inscribed in S' is self-conjugate with regard to S. If we choose for triangle of reference any triangle selfconjugate with regard to S', we have ƒ' = 0, 9' = 0, h' = 0, and ‣ becomes (bc —ƒ3) a' + (ca — g3) b' + (ab — h3) c' ; also and will vanish if we have bc=f2, ca=g', ab=h. Now bc=f" is the condition that the line x should touch S; hence vanishes if any triangle circumscribing S is self-conjugate with regard to S. In the same manner it is proved that '=0 is the condition either that it should be possible to inscribe in S a triangle self-conjugate with regard to S', or to circumscribe about S a triangle self-conjugate with regard to S. When one of these things is possible, the other is so too. A pair of conics connected by the relation =0 possesses another property. Let the point in which meet the lines joining the corresponding vertices of any triangle and of its polar triangle with respect to a conic be called the pole of either triangle with respect to that conic; and let the line joining the intersections of corresponding sides be called their axis. Then if = 0, the pole with respect to S of any triangle inscribed in S' will lie on S'; and the axis with respect to S' of any triangle circumscribing S will touch S. For eliminating x, y, z in turn between each pair of the equations ax+hy+gz=0, hx+by+fz=0, gx+fy+cz = 0, we get (gh—af) x = (hf — bg) y = (fg — ch) z, for the equations of the lines joining the vertices of the triangle xyz to the corresponding vertices of its polar triangle with respect to S. These equations may be written Fx = Gy = Hz, 1 1 1 and the coordinates of the pole of the triangle are 'H' F'G' Substituting these values in S', in which it is supposed that the coefficients a', b', c' vanish, we get 2Ff' +2Gg' + 2Hh' =0, or =0. The second part of the theorem is proved in like manner. Ex. 1. If two triangles be self-conjugate with regard to any conic S', a conic can be described passing through their six vertices; and another can be described touching their six sides (see Ex. 7, Art. 356). Let a conic be described through the three vertices of one triangle and through two of the other, which we take for x, y, z. Then, because it circumscribes the first triangle, '= 0, or a + b + c = 0 (Ex. 2, Art. 371), and, because it goes through two vertices of xyz, we have a = 0, b = 0, therefore c = 0, or the conic goes through the remaining vertex. The second part of the theorem is proved in like manner. Ex. 2. The square of the tangent drawn from the centre of a conic to the circle circumscribing any self-conjugate triangle is constant, and = a2 + b2 [M. Faure] This is merely the geometrical interpretation of the condition = 0, found (Ex. 4, Art. 371), or a2 + ß2 — p2 = a2+b2. The theorem may be otherwise stated thus: "Every circle which circumscribes a self-conjugate triangle cuts orthogonally the circle which is the locus of the intersection of tangents mutually at right angles." For the square of the radius of the latter circle is a2 + b2. Ex. 3. The centre of the circle inscribed in every self-conjugate triangle with respect to an equilateral hyperbola lies on the curve. This appears by making b2a2 in the condition ' = 0 (Ex. 4, Art. 371). Ex. 4. If the rectangle under the segments of one of the perpendiculars of the triangle formed by three tangents to a conic be constant and equal to M, the locus of the intersection of perpendiculars is the circle x2 + y2 = a2 + b2 + M. For → = = 0 (Ex. 4, Art. 371) is the condition that a triangle self-conjugate with regard to the circle can be circumscribed about S. But when a triangle is self-conjugate with regard to a circle, the intersection of perpendiculars is the centre of the circle and M is the square of the radius (Ex. 3, Art. 278). The locus of the intersection of rectangular tangents is got from this example by making M = 0. Ex. 5. If the rectangle under the segments of one of the perpendiculars of a triangle inscribed in S be constant, and IM, the locus of intersection of perpen = diculars is the conic concentric and similar with S, S= M (2+12) [Dr. Hart]. This follows in the same way from ' = 0. a2 Ex. 6. Find the locus of the intersection of perpendiculars of a triangle inscribed in one conic and circumscribed about another [Mr. Burnside]. Take for origin the centre of the latter conic, and equate the values of M found from Ex. 4 and 5; then if a', b' be the axes of the conic S in which the triangle is inscribed, the equation of a'26'2 the locus is x2 + y2 — a2 - b2 = a+b2 S. The locus is therefore a conic, whose axes are parallel to those of S, and which is a circle when S is a circle. Ex. 7. The centre of the circle circumscribing every triangle, self-conjugate with regard to a parabola, lies on the directrix. This and the next example follow from = 0 (Ex. 5, Art. 371). Ex. 8. The intersection of perpendiculars of any triangle circumscribing a parabola lies on the directrix. Ex. 9. Given the radius of the circle inscribed in a self-conjugate triangle, the locus of centre is a parabola of equal parameter with the given one. V 376. If two conics be taken arbitrarily it is in general not possible to inscribe a triangle in one which shall be circumscribed about the other; but an infinity of such triangles can be drawn if the coefficients of the conics be connected by a certain relation, which we proceed to determine. Let us suppose that such a triangle can be described, and let us take it for triangle of reference; then the equations of the two conics must be reducible to the form S = x2 + y2+z2 - 2yz - 2zx - 2xy = 0, S' = 2fyz+2gzx + 2hxy = 0. Forming then the invariants we have A=-4, =4(f+g+h), O' = − (f+g+h)", ▲'=2fgh; values which are evidently connected by the relation O2 = 44'.* * This condition was first given by Prof. Cayley (Philosophical Magazine, vol. vi. p. 99) who derived it from the theory of elliptic functions. He also proved, in the same way, that if the square root of k3▲ + k20 + k↔' + A', when expanded in powers of k, be A+ Bk + Ck2 + &c., then the conditions that it should be possible to have a polygon of n sides inscribed in U and circumscribing V, are for n = 3, 5, 7, &c. respectively C = 0, C, D C, D, E This is an equation of the kind (Art. 371) which is unaffected by any change of axes; therefore, no matter what the form in which the equations of the conics have been originally given, this relation between their coefficients must exist, if they are capable of being transformed to the forms here given. Conversely, it is easy to show, as in Ex. 1, Art. 375, that when the relation holds 2=4^O', then if we take any triangle circumscribing S, and two of whose vertices rest on S', the third must do so likewise. Ex. 1. Find the condition that two circles may be such that a triangle can be inscribed in one and circumscribed about the other. Let D2 — p2 — 7'2 = G, then the condition is (see Ex. 3, Art. 371) (G − p2)2 + 4r2 (G — go12) = 0, or (G + p2)2 = 4r2p12 ; whence D2='2 + 2rr', Euler's well known expression for the distance between the centre of the circumscribing circle and that of one of the circles which touch the three sides. Ex. 2. Find the locus of the centre of a circle of given radius, circumscribing a triangle circumscribing a conic, or inscribed in an inscribed triangle. The loci are curves of the fourth degree, except that of the centre of the circumscribing circle in the case of the parabola, which is a circle whose centre is the focus, as is otherwise evident. Ex. 3. Find the condition that a triangle may be inscribed in S' whose sides touch respectively S+ IS', S +mS', S + nS'. Let then it is evident that S + IS' is touched by x, &c. We have then A= (2 + lf + mg + nh)2 — 2lmnfgh, e= = 2 (ƒ + g + h) (2 + lf + mg + nh) + 2fgh (mn + nl + lm), 0' = − (f + g + h)2 − 2 (l + m + n) ƒgh, ▲′ = 2fgh. Whence, obviously, {0 − ▲′ (mn + nl + lm)}2 = 4 (▲ + Imn▲′) {0' + A′ (l + m + n)}, which is the required condition. 377. To find the condition that the line λx + y + vz should pass through one of the four points common to S and S'. This is, in other words, to find the tangential equation of these four points. Now we get the tangential equation of any conic of 17152 the system S+kS' by writing a+ka', &c. for a, &c. in the tangential equation of S, or Σ = (bc −ƒ2) x2 + (ca − g2) μ2 + (ab — h2) v2 − af) μv + 2 (hf − bg) vλ + 2 (fg − ch) λμ = 0. + k2E' = 0, where 2ff'') λ2 + (ca' + c'a — 2gg') μ3 -- +(ab' + a'b — 2hh') v2 + 2 (gh' + g'h — af' — a'ƒ) μv + 2 (hf' + h'ƒ — bg' — b'g) vλ + 2 (fg' + f'g — ch' — c'h) λμ. The tangential equation of the envelope of this system is there- It will be observed that =0 is the condition obtained (Art. 335), that the line λx + y + vz shall be cut harmonically by the two conics. 378. To find the equation of t e four common tangents to two conics. This is the reciprocal of the problem of the last Article, and is treated in the same way. Let & and ' be the tangential equations of two conics, then (Art. 298) ≥ + kΣ' represents tangentially a conic touched by the four tangents common to the two given conics. Forming then, by Art. 285, the trilinear equation corresponding to Σ+kΣ' = 0, we get where AS+kF+k'A'S' = 0, F = (BC' + B' C − 2FF") x* + (CA' + C'A − 2 G G′) y* +(AB' + A'B−2HH') za + 2 (GH'+G'H – AF' – A'F) yz + 2 (HF'+ H'F— BG'— B'G) zx +2 (FG' + FG - CH' — C'H) xy, |