Now, since the arc of any circle is proportional to the angle it subtends at the centre (Euc. VI. 33), and also to the radius (Art. 391), if we consider PP' as the arc of a circle, whose centre is N, the angle PNP' is measured by taking FR=FP, PFP is measured by PP' In like manner, therefore, denoting this angle by 0, PN by R, FP, F'P, by p, p', we have 2 = 1 1 R sin 0 P ρ Hence it may be inferred, that the focal chord of curvature is double b the harmonic mean between the focal radii. Substituting for b sin 0, 2a for p + p', and b' for pp', we obtain the known value b'3 R: ab. The radius of curvature of the hyperbola or parabola can be investigated by an exactly similar process. In the case of the parabola we have p' infinite, and the formula becomes I owe to Mr. Townsend the following investigation, by a different method, of the length of the focal chord of curvature: Draw any parallel QR to the tangent at P, and describe a circle through PQR meeting the focal chord PL of the conic at C. Then, by the circle PS. SC- QS. SR, and by Q the conic (Ex. 2, Art. 193) PS.SL:QS. SR :: PL: MN; therefore, whatever be the circle, SC: SL:: MN: PL; but for the circle of curvature the M P RN points S and P coincide, therefore PC: PL :: MN: PL; or, the focal chord of curvature is equal to the focal chord of the conic drawn parallel to the tangent at the point (p. 219, Ex. 4). 398. The radius of curvature of a central conic may otherwise be found thus: Let T be an indefinitely near point on the curve, QR a parallel to the tangent, meeting the normal in S; now, if a circle be described passing through P, Q, and touching PT at P, since QS is a perpendicular let fall from Q on the diameter of this circle, we have PQ=PS multiplied by the diameter; PQ* or the radius of curvature = Now, since QR is always drawn parallel to the tangent, and since PQ must ultimately coincide with the tangent, we have PQ ultimately equal to QR; but, by the property of the ellipse (if we denote CP and its conjugate by a′, b'), therefore b" : a" :: QR3 : PR. RP' (= 2a'. PR), QR" = 26'2. PR b' PR Hence the radius of curvatures. Now, no matter how small PR, PS are taken, we have, by similar triangles, their PR CP a ratio Hence radius of curvature = PS CT Ρ It is not difficult to prove that at the intersection of two confocal conics the centre of curvature of either is the pole with respect to the other of the tangent to the former at the intersection. 398 (a). If we consider the circle circumscribing the triangle formed by two tangents to a curve and their chord, it is evident geometrically, that its diameter is the line joining the intersection of tangents to the intersection of the corresponding normals. Hence, in the limit, the diameter of the circle circumscribing the triangle formed by two consecutive tangents and their chord is the radius of curvature; that is to say, the radius of the circle here considered is half the radius of curvature (Compare Art. 262, Ex. 4). 399. If two tangents be drawn to an ellipse from any point of a confocal ellipse, the excess of the sum of these two tangents over the arc intercepted between them is constant.* For, take an indefinitely near point T", and let fall the perpendiculars TR, T'S, then (see fig.) PT=PR=PP' + P'R (for P'R may be considered as the continuation of the line PP') in like manner TT' Hence (PT+ TQ) − (P'T'+ T' Q') = PP' - Q Q' = PQ-P'Q'. COR. The same theorem will be true of any two curves which possess the property that two tangents TP, TQ to the inner one always make equal angles with the tangent TT' to the outer. 400. If two tangents be drawn to an ellipse from any point of a confocal hyperbola, the difference of the arcs PK, QK is equal to the difference of the tangents TP, TQ.† For it appears, precisely as before, that the excess of T'P'-P'K over TP-PK=T'R, and that the excess of T'Q-QK over TQ-QK is T'S, which is equal to T'R, since (Art.189) TT' bisects the angle RT'S. The difference, therefore, between the excess of TP over PK, and that of TQ over QK is constant; but in the particular case where T P/P F K R S F' A * This beautiful theorem was discovered by Bishop Graves. See his Translation of Chasles's Memoirs on Cones and Spherical Conics, p. 77. †This extension of the preceding theorem was discovered by Mr. Mac Cullagh, Dublin Exam. Papers, 1841, p. 41; 1842, pp. 68, 83. M. Chasles afterwards independently noticed the same extension of Bishop Graves's theorem. Comptes Rendus, October, 1843, tom. XVII. p. 838. ССС. coincides with K, both these excesses and consequently their difference vanish; in every case, therefore, TP-PK = TQ - QK. COR. Fagnani's theorem, "That an elliptic quadrant can be so divided, that the difference of its parts may be equal to the difference of the semi-axes," follows immediately from this Article, since we have only to draw tangents at the extremities of the axes, and through their intersection to draw a hyperbola confocal with the given ellipse. The coordinates of the points where it meets the ellipse are found to be 401. If a polygon circumscribe a conic, and if all the vertices but one move on confocal conics, the locus of the remaining vertex will be a confocal conic. In the first place, we assert that if the vertex T of an angle PTQ circumscribing a conic, move on a confocal conic (see fig., Art. 399); and if we denote by a, b, the diameters parallel to TP, TQ; and by a, B, the angles TPT', TQ'T', made by each of the sides of the angle with its consecutive position, then aa=bB. For (Art. 399) TR= T'S; but TR = TP.a;T'S = T'Q'.ẞ, and (Art. 149) TP and TQ are proportional to the diameters to which they are parallel. Conversely, if aa=bß, T moves on a confocal conic. For by reversing the steps of the proof we prove that TR=T'S; hence that TT' makes equal angles with TP, TQ, and therefore coincides with the tangent to the confocal conic through T; and therefore that T' lies on that conic. If, then, the diameters parallel to the sides of the polygon be a, b, c, &c., that parallel to the last side being d, we have aa = bß, because the first vertex moves on a confocal conic; in like manner bß=cy, and so on until we find aa=dd, which shows that the last vertex moves on a confocal conic.* * This proof is taken from a paper by Dr. Hart; Cambridge and Dublin Mathematical Journal, vol. IV. 193. NOTES. PASCAL'S THEOREM, Art. 267. M. STEINER was the first who (in Gergonne's Annales) directed the attention of geometers to the complete figure obtained by joining in every possible way six points on a conic. M. Steiner's theorems were corrected and extended by M. Plücker (Crelle's Journal, vol. v. p. 274), and the subject has been more recently investigated by Messrs. Cayley and Kirkman, the latter of whom, in particular, has added several new theorems to those already known (see Cambridge and Dublin Mathematical Journal, vol. v. p. 185). We shall in this note give a slight sketch of the more important of these, and of the methods of obtaining them. The greater part are derived by joining the simplest principles of the theory of combinations with the following elementary theorems and their reciprocals: "If two triangles be such that the lines joining corresponding vertices meet in a point (the centre of homology of the two triangles), the intersections of corresponding sides will lie in one right line (their axis)." "If the intersections of opposite sides of three triangles be for each pair the same three points in a right line, the centres of homology of the first and second, second and third, third and first, will lie in a right line." Now let the six points on a conic be a, b, c, d, e, f, which we shall call the points P. These may be connected by fifteen right lines, ab, ac, &c., which we shall call the lines C. Each of the lines C (for example) ab is intersected by the fourteen others; by four of them in the point a, by four in the point b, and consequently by six in points distinct from the points P (for example the points (ab, cd), &c.). These we shall call the points p. There are forty-five such points; for as there are six on each of the lines C, to find the number of points p, we must multiply the number of lines C by 6, and divide by 2, since two lines C pass through every point p. If we take the sides of the hexagon in the order abcdef, Pascal's theorem is, that the three p points, (ab, de), (cd, fa), (bc, ef), lie in one right line, which we may call either the Pascal abcdef, or else we may denote as the Pascal (ab.cdef de.fa.bc' a form which we sometimes prefer, as showing more readily the three points through which the Pascal passes. Through each point p four Pascals can be drawn. Thus through (ab, de) can be drawn abcdef, abfdec, abcedf, abfedc. We then find the total number of Pascals by multiplying the number of points p by 4, and dividing by 3, since there are three points p on each Pascal. We thus obtain the number of Pascal's lines 60. We might have derived the same directly by considering the number of different ways of arranging the letters abcdef. Consider now the three triangles whose sides are ab, cd, ef, (1) de, fa, bc, (2) ef, be, ad, (3) |