The condition =0 found, Art. 377, is evidently a contravariant of the system of conics S, S'. 381. It will be found that the equation of any conic covariant with S and S' can be expressed in terms of S, S', and F; while its tangential equation can be expressed in terms of Σ, Σ', . Φ. Ex. 1. To express in terms of S, S', F the equation of the polar conic of S with respect to S'. From the nature of covariants and invariants, any relation found connecting these quantities, when the equations are referred to any axes, must remain true when the equations are transformed. We may therefore refer S and S' to their common self-conjugate triangle and write Sax2 + by2 + cz2, S' = x2 + y2 + z2. It will be found then that F = a (b + c) x2 + b (c + a) y2 + c (a + b) z2. Now since the condition that a line should touch S is bcλ2 + caμ2 + abv2 = 0, the locus of the poles with respect to S' of the tangents to S is bcx2 + cay2 + abz2 = 0. But this may be written (bc+ca + ab) (x2 + y2 + 2) = F. The locus is therefore (Ex. 1, Art. 371) OS' F. In like manner the polar conic of S' with regard to S is O'S F. Ex. 2. To express in terms of S, S', F the conic enveloped by a line cut harmonically by S and S'. The tangential equation of this conic = 0 is (b + c) x2 + (c + a) μ2 + (a + b) v2 = 0. Hence its trilinear equation is or or or (c + a) (a + b) x2 + (a + b) (b + c) y2 + (c + a) (b + c) z2 = 0, Ex. 3. To find the condition that F should break up into two right lines. It is abc (b+c) (c + a) (a + b) = 0, or abc {(a + b + c) (bc + ca + ab) — abc} = 0, ΔΔ' (ΘΘ' - ΔΔ') = 0, which is the required formula. 00' ▲▲' is also the condition that should break up into factors. This condition will be found to be satisfied in the case of two circles which cut at right angles, in which case any line through either centre is cut harmonically by the circles, and the locus of points whence tangents form a harmonic pencil also reduces to two right lines. The locus and envelope will reduce similarly if D2 = 2 (No2 + p22). Ex. 4. To reduce the equations of two conics to the forms x2 + y2 + z2 = 0, ax2 + by2 + cz2 = 0. The constants a, b, c are determined at once (Ex. 1, Art. 371) as the roots of x2 + y2 + z2 = S, ax2 + by2 + cz2 = S', a (b + c) x2 + b (c + a) y2 + c (a + b) z2 = F, we find x2, y2, z2 in terms of the known functions S, S', F. Strictly speaking, we ought to commence by dividing the two given equations by the cube root of A, since we want to reduce them to a form in which the discriminant of S shall be 1. But it will be seen that it will come to the same thing if leaving S and S'unchanged, we calculate F from the given coefficients and divide the result by A. Ex. 5. Reduce to the above form 3x2 - 6xy + 9y2 – 2x + 4y = 0, 5x2 - 14xy + 8y2 -- 6x − 2 = 0. It is convenient to begin by forming the coefficients of the tangential equation, 4, -1, 18; -3, 3, 2; 16, 19, -9; 21, 24, —– 14, A, B, &c. These are Ex. 6. To find the equation of the four tangents to S at its intersections with S'. Ans. (OS - AS')2 = 4AS (O'S — F). Ex. 7. A triangle is circumscribed to a given conic; two of its vertices move on fixed right lines λx + μy + vz, X'x + μ'y + v'z : to find the locus of the third. It was proved (Ex. 2, p. 239) that when the conic is z2. xy, and the lines ax - y, bx - y, the locus is (a + b)2 (z2 − xy) = (a - b)2 z2. Now the right-hand side is the square of the polar with regard to S of the intersection of the lines, which in general would be P= (ax+by+gz) (μv′ — μ’v) + (hx + by +ƒz) (v\' − v′λ) + (gx+fy + cz) (\μ′ − X'μ) = 0 ; and a+b=0 is the condition that the lines should be conjugate with respect to S, which in general (Art. 373) is → = 0, where ✪ = AXX' + Âμμ' + Cvv' + F' (uv' + μ'v) + G (vλ' + v'λ) + H (\μ′ + X'μ) = 0. The particular equation, found p. 239, must therefore be replaced in general by O2U+AP2 = 0. Ex. 8. To find the envelope of the base of a triangle inscribed in S and two of whose sides touch S'. Take the sides of the triangle in any position for lines of reference, and let where x and y are the lines touched by S'. Then it is obvious that kS+ S' will be touched by the third side z, and we shall show by the invariants that this is a fixed conic. We have ▲ = 2fgh, → = − (ƒ + g + h)2 − 2ƒghk, ✪' = 2 (f + g + h) (2 + hk), ▲′ =— (2 + hk)2, whence →'2 40▲ = 4▲▲, and the equation kS+S' = 0 may be written in the form (02 – 404) S + 4AA'S' = 0, which therefore denotes a fixed conic touched by the third side of the triangle. It is obvious that when 02 40▲ the third side will always touch S'. Ex. 9. To find the locus of the vertex of a triangle whose three sides touch a Iconic U and two of whose vertices move on another conic V. We have slightly altered the notation, for the convenience of being able to denote by U' and V' the results of substituting in U and V the co-ordinates of the vertex x'y'z'. The method we pursue is to form the equation of the pair of tangents to U through x'y'z'; then to form the equation of the lines joining the points where this pair of lines meets V; and, lastly, to form the condition that one of these lines (which must be the base of the triangle in question) touches V. Now if P be the polar of x'y'z', the pair of tangents is UU' - P2. In order to find the chords of intersection with V of the pair of tangents, we form the condition that UU' – P2 +λV may represent a pair of lines. This discriminant will be found to give us the following quadratic for determining A, X2A′+F+AU'V' = 0. In order to find the condition that one of these chords should touch U, we must, by Art. 372, form the discriminant of μU + (UU' – P2 + XV), and then form the condition that this considered as a function of μ should have equal roots. The discriminant is μ2▲ + μ (2U'A + λ0) + {U22A + λ (OU' + ▲ V') + λ2Ð'}, μ Substituting this value for λ in \2A' + λF' + AU'V' = 0, we get the equation of the required locus 16A3A'V-4A (4A0' – Ø2) F+ U (4A0′ — 02)2 = 0, which, as it ought to do, reduces to V when 4A0' = 02.* Ex. 10. Find the locus of the vertex of a triangle, two of whose sides touch U, and the third side aU+bV, while the two base angles move on V. It is found by the same method as the last, that the locus is one or other of the conics, touching the four common tangents of U and V, Ex. 11. To find the locus of the free vertex of a polygon, all whose sides touch U, and all whose vertices but one move on V. This is reduced to the last; for the line. joining two vertices of the polygon adjacent to that whose locus is sought, touches a conic of the form aU÷bV. It will be found if X', μ'; X", μ"; \"", "" be the values for polygons of n-1, n, and n+1 sides respectively, that X""' = μ'u", μ"" = A'X'X" (aμ" — A′ẞ\"). In the case of the triangle we have X'a, μ' A'ß; in the case of the quadrilateral \” = ß2, μ” = a (4Aα +260), and from these we can find, step by step, the values for every other polygon. (See Philosophical Magazine, Vol. XIII., p. 337). = α, Ex. 12. The triangle formed by the polars of middle points of sides of a given triangle with regard to any inscribed conic has a constant area [M. Faure]. Ex. 13. Find the condition that if the points in which a conic meets the sides of the triangle of reference be joined to the opposite vertices, the joining lines shall form two sets of three each meeting in a point. Ans. abc — 2fgh — af2 — bg2 — ch2 = 0.. 382. The theory of covariants and invariants enables us readily to recognize the equivalents in trilinear co-ordinates of certain well-known formulæ in Cartesian. Since the general expression for a line passing through one of the imaginary circular points at infinity is x+y√(−1)+c, the condition that * The reader will find (Quarterly Journal of Mathematics, Vol. I., p. 344) a discussion by Mr. Cayley of the problem to find the locus of vertex of a triangle circumscribing a conic S, and whose base angles move on given curves. When the curves are both conics, the locus is of the eighth degree, and touches S at the points where it is met by the polars with regard to S of the intersections of the two conics. X X λx+μy +v should pass through one of these points is λ2 + μ2 = 0. In other words, this is the tangential equation of these points. If then Σ=0 be the tangential equation of a conic, we may form the discriminant of Σ+ k (λ2+μ2). Now it follows from Arts. 285, 286, that the discriminant in general of Σ + kΣ' is A2+kAO' + k2A'0 + k3A22. But the discriminant of Σ+k (x2 + μ2) is easily found to be ▲*+ k▲ (a + b) + k2 (ab — h3). If, then, in any system of co-ordinates we form the invariants of any conic and the pair of circular points, →'=0 is the condition that the curve should be an equilateral hyperbola, and → = 0 that it should be a parabola. The condition (a + b)* = 4 (ab — h3), or (a−b)2+4h2 = 0, must be satisfied if the conic pass through either circular point; and it cannot be satisfied by real values except the conic pass through both, when a=b, h=0. Now the condition λ2+μ2=0* implies (Art. 34) that the length of the perpendicular let fall from any point on any line passing through one of the circular points is always infinite. The equivalent condition in trilinear coordinates is therefore got by equating to nothing the denominator in the expression for the length of a perpendicular (Art. 61). The general tangential equation of the circular points is therefore λ2 +μ3 + v3 − 2 μv cos A - 2vλ cos B-2λμ cos C=0. Forming then the → and e' of the system found by combining this with any conic, we find that the condition for an equilateral hyperbola, '=0, is a+b+c-2f cos A-2g cos B-2h cos C=0; while the condition for a parabola, 0=0, is A sin A+B sin B+C sin' C + 2F sin B sin C +2G sin C sin A+ 2H sin A sin B = 0. *This condition also implies (Art. 25) that every line drawn through one of these two points is perpendicular to itself. This accounts for some apparently irrelevant factors which appear in the equations of certain loci. Thus if we look for the equation of the foot of the perpendicular on any tangent from a focus aß, (x − a)2 + (y− ß)2 will appear as a factor in the locus. For the perpendicular from the focus on either tangent through it coincides with the tangent itself. This tangent therefore is part of the locus. The condition that the curve should pass through either circular point is "2=40, which can in various ways be resolved into a sum of squares. 383. If we are given a conic and a pair of points, the covariant F of the system denotes the locus of a point such. that the pair of tangents through it to the conic are harmonically conjugate with the lines to the given pair of points. When the pair of points is the pair of circular points at infinity, F denotes the locus of the intersection of tangents at right angles. Now, referring to the value of F, given Art. 378, it is easy to see that when the second conic reduces to λ2+μ2; that is, when A'=B'=1, and all the other coefficients of the tangential of the second conic vanish, F is С (x2 + y2) − 2 Gx−2Fy+A+B=0, which is, therefore, the general Cartesian equation of the locus of intersection of rectangular tangents. (See p. 258). When the curve is a parabola C=0, and the equation of the directrix is therefore 2 (Gx + Fy) = A + B. The corresponding trilinear equation found in the same way is (B+C+2F cos A) x2 + ( C+ A + 2 G cos B) y2+(A+B+2H cos C) z2 +2 (A cos A-F-G cos C-H cos B) yz +2 (B cos B-G-H cos A-F cos C) zx. +2(C cos C – H-F cos B-G cos A) xy = 0. It may be shown, as in Art. 128, that this represents a circle, by throwing it into the form (x sin4+y sinB+ z sin C') (B+C+2F A+B+2H cos C + sin C x) = cosA sin C+A+2 G cosB sin B Ꮎ sin▲ sinBsin C (yzsinA+zxsin B+xy sin C'), where is the condition (Art. 382) that the curve should be a parabola. When = 0, this equation gives the equation of the directrix. 384. In general, +k' denotes a conic touching the four tangents common to Σ and '; and when k is determined so that represents a pair of points, those points are two |