or 12 12 13 00" + 18Δ Δ' Θ' - 27 Δ Δ' - ΑΔΘ" - 4Δ' Ό = 0, which is the required condition that the conics should touch. It is proved, in works on the theory of equations, that the left-hand member of the equation last written is proportional to the product of the squares of the differences of the roots of the equation in k; and that when it is positive the roots of the equation in k are all real, but that when it is negative two of these roots are imaginary. In the latter case (see Art. 282), S and S' intersect in two real and two imaginary points: in the former case, they intersect either in four real or four imaginary points. These last two cases have not been distinguished by any simple criterion. Ex. 1. To find by this method the condition that two circles should touch. Forming the condition that the reduced equation (Ex. 3, Art. 371), ‚·2+ (r2+r22 — D2) k + r22k2 = 0, should have equal roots, we get r2 + p′2 − D2 = ± 2rr' ; D = r±r' as is geometrically evident. Ex. 2. Find the locus of the centre of a circle of constant radius touching a given conic. We have only to write for A, A', 0, e' in the equation of this article, the values Ex. 4 and 5, Art. 371; and to consider a, ẞ as the running co-ordinates. The locus is in general a curve of the eighth degree, but reduces to the sixth in the case of the parabola. This curve is the same which we should find by measuring from the curve on each normal, a constant length, equal to r. It is sometimes called the curve parallel to the given conic. Its evolute is the same as that of the conic. The following are the equations of the parallel curves given at full length, which may also be regarded as equations giving the length of the normal distances from any point to the curve. The parallel to the parabola is p6 — (3y2 + x2 + 8mx − 8m2) r1 + {3y2 + y2 (2x2 - 2mx + 20m2) +8mx3 + 8m2x2 - 32m3x + 16m1} r2 — (y2 — 4mx)2 {y2 + (x − m)2} = 0. The parallel to the ellipse is · 2c2p6 {c2 (a2 + b2) + (a2 − 2b2) x2 + (2a2 — b2) y2} + p$ {c1 (a1 + 4a2b2 + b1) − 2c2 (a1 — a2b2 + 3b1) x2 + 2c2 (3a1 — a2b2 + b1) y2 + (a1 — 6a2b2 + 6b1) x2 + (6a1 — 6a2b2 + b1) y1 + (6aa — 10a2b2 + 6b1) x2y2} + p2 {— 2a2b2c1 (a2 + b2) + 2c2x2 (3a1 — a2b2 + b1) — 2c2y2 (a1 — a2b2 + 3b1) →b2x2 (6a2 – 10a2b2 + 6b1) — a2y1 (6aa — 10a2b2 + 6b1) + x2y2 (4a6 — 6aab2 — 6a2b1 + 4b6) +2b2 (a2 — 2b2) x6 − 2 (aa — a2b2 +3b1) x1y2 — 2 (3a1 — a2b2 + b1) x2y1 + 2a2 (b2 — 2a2) yo} + (b2x2 + a2y2 — a2b2)2 {(x − c)2 + y2} {(x + c)2 + y2} = 0. Thus the locus of a point is a conic, if the sum of squares of its normal distances to the curve be given. If we form the condition that the equation in r2 should have equal roots, we get the squares of the axes multiplied by the cube of the evolute. If we make 0, we find the foci appearing as points whose normal distance to the curve vanishes. This is to be accounted for by remembering that the distance from the origin vanishes of any point on either of the lines x2 + y2 = 0. Ex. 3. To find the equation of the evolute of an ellipse. Since two of the normals coincide which can be drawn through every point on the evolute, we have only to express the condition that in Ex., Art. 370 the curves S and S' touch. Now when the term 2 is absent from an equation, the condition that Ak3 + O'k + A′ should have equal roots, reduces to 27AA'2 + 40′30. The equation of the evolute is therefore (a2x2 + b2y2 — c1)3 + 27a2b2c1x2y2 = 0. (See Art. 248). Ex. 4. To find the equation of the evolute of a parabola. We have here S = y2 - 4mx, S' = 2xy + 2 (2m − x′) y — 4my', and the equation of the evolute is 27my2 = 4 (x - 2m)3. It is to be observed, that the intersections of S and S' include not only the feet of the three normals which can be drawn through any point, but also the point at infinity on y. And the six chords of intersection of S and S', consist of three chords joining the feet of the normals, and three parallels to the axis through these feet. Consequently the method used (Ex., Art. 370) is not the simplest for solving the corresponding problem in the case of the parabola. We get thus the equation found (Ex. 12, p. 203), but multiplied by the factor 4m (2my + y'x — 2my') — yˆ3⁄43. 373. If S' break up into two right lines we have ▲'=0, and we proceed to examine the meaning in this case of ✪ and e'. Let us suppose the two right lines to be x and y; and, by the principles already laid down, any property of the invariants, true when the lines of reference are so chosen, will be true in general. The discriminant of S+2hxy is got by writing h+k for h in A, and is A+2k (fg-ch) - ck. Now the coefficient of k vanishes when c=0, that is, when the point xy lies on the curve S. The coefficient of k vanishes when fg=ch; that is (see Ex. 3, p. 204), when the lines x and y are conjugate with respect to S. Thus, then, when S' represents two right lines, ▲′ vanishes; '=0 represents the condition that the intersection of the two lines should lie on S; and ● = 0 is the condition that the two lines should be conjugate with respect to S.. The condition that A+ ok+o'k should be a perfect square is 02 = 440', which, according to the last Article, is the condition that either of the two lines represented by S' should touch S. This is easily verified in the example chosen, where 02 – 440′ is found to be equal to (bc-ƒ") (ca — g3). Ex. 1. Given five conics S1, S2, &c. it is of course possible in an infinity of ways to determine the constants 11, 12, &c. so that 11S1 + 12S2 + 13S3 + 14S4 + 15S5 may be either a perfect square L2, or the product of two lines MN: prove that the lines L all touch a fixed conic V, and that the lines M, N are conjugate with regard to V. We can determine V so that the invariant → shall vanish for V and each of the five conics, since we have five equations of the form Aa1 + Bb1+Cc1 + 2Ff1+2Gg, + 2Hh1 = 0, which are sufficient to determine the mutual ratios of A, B, &c., the coefficients in the tangential equation of V. Now if we have separately Aa, + &c. = 0, Aα, + &c. = 0, Aa3+ &c. = 0, &c., we have plainly also ▲ (la1 + 11⁄2ɑ2 + 13ɑ3 + 14ɑ4 + 15α5) + &c. = 0 ; that is to say, O vanishes for V and every conic of the system 11S1 + 12S2 + 13S3 + 14S4 + 15 S5, whence by this article the theorem stated immediately follows. If the line M be given, N passes through a fixed point; namely, the pole of M with respect to V. Ex. 2. If six lines x, y, z, u, v, w all touch the same conic, the squares are connected by a linear relation 4x2 + l2y2 + 1322 + 1⁄4u2 + 15v2 + low2 = 0. This is a particular case of the last example; but may be also proved as follows: Write down the conditions, Art. 151, that the six lines should touch a conic, and eliminate the unknown quantities A, B, &c., and the condition that the lines should touch the same conic is found to be the vanishing of the determinant But this is also the condition that the squares should be connected by a linear relation. Ex. 3. If we are only given four conics S1, S2, S3, S4, and seek to determine V, as in Ex. 1, so that ✪ shall vanish, then, since we have only four conditions, one of the tangential coefficients A, &c. remains indeterminate, but we can determine all the rest in terms of that; so that the tangential equation of V is of the form Σ+k' = 0, or V touches four fixed lines. We shall afterwards show directly that in four ways we can determine the constants so that S1 + 12S2 + 13S3 + 184 may be a perfect square. It is easy to see (by taking for M the line at infinity) that if M be a given line it is a definite problem admitting of but one solution to determine the constants, so that S,&c. shall be of the form MN. And Ex. 1 shows that N is the locus of the pole of M with regard to V. Compare Ex. 8, p. 205. 374. To find the equation of the pair of tangents at the points where S is cut by any line λx + μy + 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 o' also. And if we denote by Σ the quantity Ax2 + Bμ2 + Cv2 + 2Fμv+2 Gvλ + 2Hλμ; the equation to determine k has two roots = 0, the third root being given by the equation k▲ += 0. The equation of the pair of tangents is therefore ES=▲ (λx + μy + vz)2. It is plain that when λx +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 ax2+by+cz" (Art. 258). We have therefore f=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, g′ =0, h'=0, and → becomes (bc —ƒ2) a' + (ca — g3) b' + (ab — h2) c' ; and will vanish if we have be=f, ca=g, ab=h'. Now bc=f2 is the condition that the line x should touch S; hence, → also 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=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 we get ax+hy+gz=0, hx+by+fz=0, gx+fy+cz = 0, 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 and the co-ordinates 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=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, p. 310). 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, O' = 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 b2 = a2 in the condition '0 (Art. 371, Ex. 4). 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, p. 243). 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 M, the locus of intersection of perpen diculars is the conic concentric and similar with S, S= M (+) [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 = S. The locus is therefore a conic, whose axes a'2+ b22 are parallel to those of S, and which is a circle when S is a circle, UU |