11. After a little practice, the process of reciprocating a given theorem will be found to consist simply in writing straight line" for "point," "join" for "intersect," "locus for "envelope," &c., and vice versa. The word "conic" will of course remain unaltered. By reciprocating Paschal's Theorem (given in Art. 12, Chap. III.), we obtain Brianchon's Theorem, which asserts that "If a hexagon be described about a conic section, the three diagonals will intersect in a point*.* * It may be well to append an independent proof of this important theorem. Take three sides of the hexagon as lines of reference, and let the equations of the other three be a+m1ß+n1y=0, l2a+B+n,y=0, l ̧a+m2ß+y=0. Let the equation of the conic be whence m1 n1 The line passing through the intersections of ẞ=0 with (l ̧, m3, 1) and of y=0 with (12, 1, n) is represented by the equation The student will find it useful to transform, by the method of reciprocal polars, the special cases of Pascal's Theorem, given in Art. 13, Chap. III.; and to obtain a geometrical construction by which when five tangents to a conic are given, their points of contact may be found. 13. The anharmonic ratio of the pencil formed by four intersecting straight lines is the same as that of the range formed by their poles. This may be proved as follows. Let OP, OQ, OR, OS be the four straight lines, P', Q', R', S' their poles, which will lie in a straight line, the polar Q'R' S Fig. 18. of 0; let P, Q, R, S be the points in which the pencil is cut by the transversal P'Q'R'S. Let this transversal cut the conic in K, K ̧. Bisect 2 K,K, in V. Then, since PKPK, is divided harmonically in P, K1, P', K, (Art. 21, Chap. II.), it follows that whence 2 PK1.P'K ̧=PK.P'K ̧ (VP – VK ̧) (VK2+ VP')=(VP+VK ̧) (VK ̧− VP'), which since VK1 = VK,, reduces to = VQ.VQ = VR. VR' = VS. VS'. Hence the eight points P, Q, R, S, P', Q', R', S' are in an involution, of which K1, K, are the foci, and therefore (Art. 27, Chap. I.) 2 {0. PQRS} = [P'Q'R'S']. 14. In Art. 13, Chap. I. we saw that the condition that the three points (11, m1, n1), (1,, m, n), (l ̧, m ̧, n ̧) shall lie in the same straight line is identical with the condition that the three straight lines (,, m,, n1), (l, m, n), (3, Mg, Ng) shall intersect in the same point. Now these several points and lines are respectively the poles and polars of each other, with respect to the imaginary conic Thus the theory of reciprocal polars explains the fact that the condition for three points lying in a straight line is identical with that for three straight lines insersecting in a point. It also explains the identity of conditions noticed in Chap. II. Arts. 7 and 9. For the reciprocal of the conic + β + γ - 2μνβγ - 2νλγα - 2λμαβ = 0 .. (1), with respect to a2 + ß2 + y2 = 0, will be found to be λβη + μγα + ναβ = 0.... (2). And the polar of (f, g, h) is fa+gB + hy = 0. Hence if the line fa+gB+hy=0 touch (1), the point (f, g, h) lies in (2), giving for the condition of tangency And if the line fa+gB+hy=0 touch (2), the point (f, g, h) lies in (1), giving for the condition of tangency in that case \2ƒ2 + μ3g2 + v3h2 — 2μvgh — 2vλhƒ — 2λμfg = 0. These conditions of tangency are identical with those already investigated. Again, every parabola touches the line at infinity. Now the co-ordinates of the pole of this line are proportional to a, b, c. Hence, if the conic, represented by the general equation of the second degree, be a parabola, the point (a, b, c) must lie in the reciprocal conic. This gives, as the condition for a parabola, Ua2 + Vb2 + Wc2 + 2 Ubc + 2 V'ca + 2 W'ab = 0, the same as that already investigated. 15. PROP. Any straight line drawn through a given plane A is divided harmonically by any conic section, and the polar of A with respect to it. This proposition may be proved as follows. Let the straight line cut the curve in P and Q, and the polar of A in B. Let C be the polar of the straight line, and let ABC be the triangle of reference. The conic will be self-conjugate with respect to ABC, and will therefore be represented by the equation ua2 + vẞ2 + wy2 = 0. Hence the lines CP, CQ, which are tangents to the conic, are represented by the equation ua2 + vß2 = 0, and therefore form an harmonic pencil with CA, CB. 16. The straight line CB may be regarded as the polar of A with respect to the locus made up of the two straight lines CP, CQ. For the values of ẞ and y at A being 0, 0, and the equation of CP, CQ being ua* + v2 = 0, we get for the equation of its polar, a = 0, that is, the polar is the line AB. 17. If four straight lines form an harmonic pencil, either pair will be its own polar reciprocal with respect to the other. For, adapting the equation of Art. 8, to the case of two variables only, we get for the polar reciprocal of aß = 0, with respect to ua2 + vß2 = 0, the following equation, And, conversely, for that of ua+vß*0 with respect to aß = 0, or ua+vẞ2 = 0, in either case reproducing the reciprocated Hence the proposition is proved. curve. 18. We may hence deduce the condition that two pairs of straight lines may form an harmonic pencil. First let them all intersect in A, and the equations of the two pairs be ua2 + vß2 + 2w'aß = 0....... The polar reciprocal of (1) with respect to (2) is pa +r'ß, r'a+qB (1), (2). 0, pa + r'ß, |