Solving these equations, we get equations of the form λ=Lλ'+L'μ'+L"v', μ = MX'+M'μ'+M"v, v=Nλ'+ N'μ'+N"v'. If then we put these values into the condition as first obtained in terms of λ, μ, v, we get the condition in terms of X', u', v', which can only differ by a constant multiplier from the condition as obtained by the other method. Functions of the class here considered are called contravariants. Contravariants are like covariants in this: that any contra variant equation, as for example, the tangential equation of a conic (bc-f2) x2 + &c. = 0 can be transformed by linear substitution into the equation of like form (b'c' -ƒ") λ"2 + &c. = 0, formed with the coefficients of the transformed trilinear equation of the conic. But they differ in that λ, μ, v are not transformed by the same rule as x, y, z; that is, by writing for λ, λ + mμ + nv, &c., but by the different rule explained above. 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 terins of 2, 2′, Þ. 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 S = ax2 + by2 + cz2, S′ = x2 + y2 + z2. It will be found then that Fa (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 bex2 + cay2 + abz2 = 0. written (bc+ca + ab) (x2 + y2 + z2) = F. The locus is therefore 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 (b + c) λ2 + (c + a) μ2 + (a + b) v2 = 0. Hence its trilinear equation is or or or But this may be (Ex. 1, Art. 371) = 0 is (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) — abe} = 0, which is the required formula. 00' AA' 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 (r2 + go′2). 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 And if we then solve the equations x2 + y2 + z2 = 8, 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 equations A, B, &c. These are 4, 1, 18; — 3, 3, −2; — 16, — 19, -9; 21, 24, — 14. We have then = A = 9, 054, 0′ = 99, ▲' = − 54, whence a, b, c are 1, 2, 3. We next calculate F which is = 2S+3S F, Z2 = (x + y + 1)2. Ex. 6. To find the equation of the four tangents to S at its intersections with S'. Ans. (OSAS′)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, Art. 272) that when the conic is z2 - xy, and the lines ax - y, bx-y, the locus is (a + b)2 (22 — xy) = (a - b) 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+hy+gz) (uv′ — μ'v) + (hx + by + ƒz) (v\' − v'λ) + (gx +ƒy + cz) (Xμ' — 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 ✪ = Aλλ' + Bμμ' + Cvv' + F' (uv' + μ'v) + G (vλ' + v′λ) + H (λμ' + X'μ) = 0. The particular equation, found Art. 272, 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 S = 2 (fyz+gzx + hxy), S': = x2 + y2 + z2 — 2yz - 2zx - 2xy - 2hkxy, where & and are the lines touched by S'. Then it is obvious that kS+ S' will 1 for the length of a perpendicular (Art. 61). The general tangential equation of the circular points is therefore x2++-2μv cos A-2vλ cos B-2 cos C=0. Forming then the and O' 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, 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. The condition that the curve should pass through either circular point is 0=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 X+; that is, when A'B'=1, and all the other coefficients of the tangential of the second conic vanish, Fis C(x2 + y) − 2 Gx−2Fy + A + B = 0, which is, therefore, the general Cartesian equation of the locus of intersection of rectangular tangents. (See Art. 294, Ex.). When the curve is a parabola C=0, and the equation of the directrix is therefore 2 (Ge+Fy)=A+B The corresponding trilinear equation found in the same way is (B+C+2Fcos A)x+(C+A+2G cos By2+(A+B+ 2H cos C) = +2(A cos A-F-G cos C-II cos B) yz +2(B cos B-G-Hcos A-F cos C)zz +2/C cos C-H-Fcos B-G cos A) ry=0, find, step by step, the values for every other polygon. (See Philosophical Magazine, vol. XIII. p. 337). 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 coordinates 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 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 20 be the tangential equation of a conic, we may form the discriminant of 2+k (+μ). Now it follows from Arts. 285, 286, that the discriminant in general of Σ + kΣ' is = ▲2+ k^®' + k2A'© + k3A”. But the discriminant of Σ + k (λ*+μ") is easily found to be ▲2 + k▲ (a + b) + k2 (ab — h3). If, then, in any system of coordinates 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)" + 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 *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. for the length of a perpendicular (Art. 61). The general tangential equation of the circular points is therefore x2 + μ3 + v2 − 2μv cos A - 2vλ cos B-2λμ cos C=0. Forming then the this with any conic, hyperbola '0, is and ' of the system found by combining we find that the condition for an equilateral a+b+c-2f cos A-2g cos B-2h cos C=0; while the condition for a parabola =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. The condition that the curve should pass through either circular point is "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 λ+ "; that is, when A' = B' = 1, and all the other coefficients of the tangential of the second conic vanish, F is C (x2 + y2) − 2 Gx − 2Fy + A + B =0, which is, therefore, the general Cartesian equation of the locus of intersection of rectangular tangents. (See Art. 294, Ex.). 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+2Fcos A) x2 + ( C + A + 2 G cos B) y2 + (A+B+2H cos C) ≈* +2(A cos A - F - G cos C-II cos B) yz +2 (B cos B-G - H cos A - F cos C) zx |