2 ANHARMONIC PROPERTIES. 113 KK in V. Then, since PK, PK, is divided harmonically in P, K1, P', K2 (Art. 21, Chap. II.), it follows that whence PK1.P'K1=PK.P'K, (VP – VK,) (VK,+VP)=(VP+VK,) (VK, – VP), which since VK1= VK2, reduces to 14. In Art. 13, Chap. I. we saw that the condition that the three points (11, m1, n1), (12, m,, n), (1 ̧, m., n.) shall lie in the same straight line is identical with the condition that the three straight lines (, m1, n1) (12, m2, n2), (13, M., n3) 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 a2 + B2 + y2 = 0. 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 intersecting 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 - 2μνβη - 2νλγα - 2λμαβ = 0 ...... (1), with respect to will be found to be a2 + ß2 + y2 = 0, λβγ + 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 (2), giving for the condition of tangency in that case X3ƒa + μ3g2 + v3h2 — 2μvgh — 2vλhf — 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 U'bc + 2 V'ca + 2 W'ab = 0, the same as that already investigated. 15. PROP. Any straight line drawn through a given point 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. ANHARMONIC PROPERTIES. 115 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+v80, 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 + vß2 = 0 with respect to aẞ=0, or ua2 + vß2 = 0, in either case reproducing the reciprocated curve. Hence the proposition is proved. 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+qß (1), (2). 0, r'a + qß, or u (r'a+qß)2 + v (pa + v'ß)2 — 2w' (r'a +qß) (px + r'ß) = 0. At the point of intersection of the line a+k,ß =0, with Y = 0, we have Taking the polar of this with respect to the curve (2) we get (pk ̧−r') a + (r'k ̧ − q) B = 0. If this be identical with a+k,ß=0, we get pk. — r' = "'k, — 9, or pk ̧k, −2v′ (k ̧+k2) + q = 0; the required condition. .. pv - 2r'w' + qu = 0, The symmetry of this equation shews that (2) is also its own polar reciprocal with respect to (1), as ought to be the case. 19. If the point of intersection of the four straight lines do not coincide with one of the angular points of the triangle of reference, we have then only to express the condition that the range formed by their intersection with any one of its sides, y = 0, for instance, be an harmonic range, If this be the case, the pencil formed by joining these four points with C will be an harmonic pencil, and we shall have, as before, pv-2r'w' + qu=0. RECIPROCATION WITH RESPECT TO A CIRCLE. 117 20. We next proceed to consider the results to be deduced from the theory of reciprocal polars, when the auxiliary conic is a circle. It is here that the utility of the theory is most apparent, as we are thus enabled to transform metrical theorems, i. e. theorems relating to the magnitudes of lines and angles. We know that, if PQ be the polar of a point T with respect to a circle, of which the centre is S and radius k, then ST will be perpendicular to PQ. Let ST cut PQ in V. Then ST. SV=k. Hence the pole of any line is at a distance from the centre of the auxiliary circle inversely proportional to the distance of the line. And conversely, the polar of any point is at a distance from the centre of the auxiliary circle, inversely proportional to the distance of the point itself. 21. If TX, TY be any two indefinite straight lines, P, Q their poles, then, since SP is perpendicular to TX, SQ to TY, it follows that the angle PSQ is equal to the angle XTY or its supplement, as the case may be. Hence, the angle included between any two straight lines is equal to the angle subtended at the centre of the auxiliary circle by the straight line joining their poles, or to its supplement. 22. From what has been said in Art. 15, and the earlier articles of this chapter, it will appear that to find the polar reciprocal of a given curve with respect to a circle, we may proceed by either of the following two methods. First. Draw a tangent to the curve, and from S, the centre of the auxiliary circle, draw SY perpendicular to the tangent, and on SY, produced if necessary, take a point Q, such that SQ.SY. The locus of Q will be the required polar reciprocal. Secondly. Take a point P on the curve, and join SP; on SP, produced if necessary, take a point Z, such that SP. SZ=k2. Through Z draw a straight line perpendicular to SP. The envelope of this line will be the required polar reciprocal. |