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(ƒ + g + h) (2 + hk), ▲′ = − (2 + hk)2, whence "- 40A′ = 4AA'k, and the equation kS+ S'=0 may be written in the form (0'2 – 40▲′) S +4▲▲'S' = 0, which therefore denotes a fixed conic touched by the third side of the triangle. It is obvious that when '2 = 40A' 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 coordinates 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 + XV may represent a pair of lines. This discriminant will be found to give us the following quadratic for determining A, λ2A'+λF'+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 + λV), 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) + {U'2A + λ (OU' + ▲ V') + X20'}, and the condition for equal roots gives Substituting this value for λ in λ2A′+\F' + AU'V', we get the equation of the required locus 16A3A'V – 4A (4A0′ – 02) F+ U (4A0′ − −2)2 = 0, which, as it ought to do, reduces to V when 40′ = 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, ΔΔ'λεν + λμΕ + μ20 = 0, where Aμ is given by the quadratic where a (ab — ẞa) λ2 + a (4▲a + 20b) λμ — b2μ2 = 0, 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"; \"", μ"" be the values for polygons of n- -1, n, and n+1 sides respectively, that "" = μ'μ"2, μ"”' = A'\'\"′′ (aμ" — ▲′ßì”). In the case of the triangle we have λ'= a, μ' = A'ß; in the case of the quadrilateral X′′ = ß2, μ′′ = a (4▲a + 2ß0), and from these we can * The reader will find (Quarterly Journal of Mathematics, vol. I. p. 344) a discussion by Prof. 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. 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 — bg? — 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 Σ=0 be the tangential equation of a conic, we may form the discriminant of +(x+3). Now it follows from Arts. 285, 286, that the discriminant in general of Σ + kΣ' is ▲2+ k^®' + k2A'© + k3A”. 2 But the discriminant of Σ + k (λ2 + μ3) is easily found to be ▲2 + k▲ (a + b) + k2 (ab — h2). 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)2 = 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 +μ=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 + μ2 + v2 − 2μv cos A· 2vλ cos B-2λμ cos C=0. Forming then the ℗ and ' of the system found by combining this with any conic, we find that the condition for an equilateral hyperbola ' = 0, is a+b+c-2ƒ 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 λ2+ μ2; that is, when A'=B'=1, and all the other coefficients of the tangential of the second conic vanish, F is C (x2+ y3) — 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. is The corresponding trilinear equation found in the same way (B+C+2Fcos A) x2 + ( C+ A + 2 G cos B) y2 + (A+B+2H cos C) ≈2 +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-Fcos 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 A+B+2H cos C + 2)= Θ sin C = sinA sinB sin C (yzsinA+zxsin B+xysin C'), where 0 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, Σ+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 opposite vertices of the quadrilateral formed by the common tangents. In the case where ' denotes the circular points at infinity, when 2+kΣ' represents a pair of points, these points are the foci (Art. 258a). If, then, it be required to find the foci of a conic, given by a numerical equation in Cartesian coordinates, we first determine k from the quadratic (ab - h2) k2 + ▲ (a + b) k + ▲2 = 0. Then, substituting either value of k in Σ + k (X2 + μ3), it breaks up into factors (λx' + μy' + vz') (λx" + μy" + vz"); and the foci x'y' x" y" are One value of k gives the two real foci, and the other two imaginary foci. The same process is applicable to trilinear coordinates. In general, + k (λ2 + μ3) represents tangentially a conic confocal with the given one. Forming, by Art. 285, the corresponding Cartesian equation, we find that the general equation of a conic confocal with the given one is ▲ S + k { C (x2 + y2) − 2 Gx − 2 Fy + A + B} + k2 = 0. From this we can deduce that the equation of common tangents is { C (x2 + y2) − 2 Gx − 2Fy + A + B}" = 4^ S. By resolving this into a pair of factors {(x − a)2 + ( y − B)2} {(x — a')2 + (y − B')"}, we can also get a, B; a', B' the coordinates of the foci. Ex. 1. Find the foci of 2x2 - 2xy + 2y2 - 2x - 8y+11. The quadratic here is 3k2 + 4k▲ + 2 = 0, whose roots are k▲, k¤—‡a. But A==- 9. Using the value k = 3, 6λ2+21μ2 + 3v2 + 18μv + 12vλ + 30λμ + 3 (λ2 + μ2) = 3 (✯ + 2μ + v) (3λ +4μ + v), showing that the foci are 1, 2; 3, 4. The value 9 gives the imaginary foci 2 ± √(− 1), 3 F √(− 1). Ex. 2. Find the coordinates of the focus of a parabola given by a Cartesian equation. The quadratic here reduces to a simple equation, and we find that (a + b) {Aλ2 + Bμ2 + 2Fμv + 2Gvλ + 2Hλμ} − ▲ (\2 + μ2); is resolvable into factors. But these evidently must be (a + b) (2Gλ +2Fμ) and (a + b) A - ▲ 2+ (a + b) B-A μ + ν. The first factor gives the infinitely distant focus, and shows that the axis of the curve is parallel to Fx - Gy. The second factor shows that the coordinates of the focus are the coefficients of X and μ in that factor. Ex. 3. Find the coordinates of the focus of a parabola given by the trilinear equation. The equation which represents the pair of faci is ΘΕΣ = ▲ (12 + μ2 + v2 — 2μv cos A - 2vλ cos В - 2λμ cos C). But the coordinates of the infinitely distant focus are known, from Art. 293, since it is the pole of the line at infinity. Hence those of the finite focus are A sin A+ H sin B + G sin C' H sin A + B sin B + F sin C' O'C-A G sin A+Fsin B + C sin C' 385. The condition (Art. 61) that two lines should be mutually perpendicular, λλ' + μμ' + vv' − (μv' + μ'v) cos A − (vλ' + v′λ) cos B is easily seen to be the same as the condition (Art. 293) that the lines should be conjugate with respect to λ2 + μ2 + v2 − 2μv cos A - 2vλ cos B-2λμ cos C=0. The relation, then, between two mutually perpendicular lines is a particular case of the relation between two lines conjugate with regard to a fixed conic. Thus, the theorem that the three perpendiculars of a triangle meet in a point is a particular case of the theorem that the lines meet in a point which join the corresponding vertices of two triangles conjugate with respect to a fixed conic, &c. It is proved (Geometry of Three Dimensions, Chap. IX.) that, in spherical geometry, the two imaginary circular points at infinity are replaced by a fixed |