140 



NATURE 



[Junes, 1890 



Sj, intersecting u and u-^ in k and k-^ ; and project k from S and 

 /C'l from Si by rays meeting in K, a point on the required curve. 



Proof. — Newton has solved this problem for a particular 

 case, of which we will now give a short account, and thence 

 deduce the more general modern method just described. 



Case I. — Let ABCD (Fig. 2) be a quadrangle inscribed in an 

 ellipse, and P a point on the ellipse outside of the quadrangle ; 

 then, if PS and PQ be two chords meeting the sides of the 

 quadrangle in S and T, R and Q respectively, the ratio 

 PR. PQ 

 PS . PT 

 -will be constant for all positions of P. For, in the first instance, 

 let_PR and PQ be parallel to the side AC, PS and PT parallel 



or, by subtraction — 



PQ^ RR' ^ PQ/PR - RR 

 PS . TT' PSVPT 



TT'; 



wherefore — 



PQ . P R' 

 PS . PT' 



PR' 

 PR 



PT' 

 PT' 



(5) 



Fig. 2. 



to the adjacent side AB, whilst the opposite sides AC and BD 

 are parallel to each other. Owing to the parallelism of the sides 

 AC and BD, a line bisecting those sides will be a diameter of 

 the ellipse, bisecting the line RQ in O. The line PO will be 

 the ordinate of P parallel to the axis conjugate of this diameter. 

 Now produce PO to K, making OK equal to OP ; then K 

 will be a point on the ellipse, wherefore — 



PQ . QK (xz^- 

 — ^ — >-.v, = — , a constant ratio. 

 AQ . QB \xy)' 



But 



PR = QK, PS = AQ, PT = QB'; 

 therefore — 



PQ . PR PQ QK , , 



PS7"PT =AQ-rQB'"^°"''''^'- 

 Case II.— When BD^ and AC are not parallel, draw BD 

 parallel to AC, meeting the conic in D and the line PST in T. 

 Join CD, intersecting PQ in R and D^N, a line parallel to PQ, 

 in M. Then, by sim ilar triangles — 



BT or PQ ^ DjN. 



TT' BN ' ^^' 



also — 



R'R _ DjM , 



AQ or PS AN ' ^'^' 



whence, by multiplication — 



PQ ■ RR' ^ DjN ■ D^M , , 



PS . TT' AN . BN ^^' 



But, by Case I., since Dj is a point on the ellipse and similarly 

 situated with respect to M and N as P is with respect to R and 

 •Q, we have — 



PQ ■ PR ^ DjN . D,M . , 



PS . PT AN . BN ^^^ 



Hence, from equations (3) and (4) — 



PQ ■ RR' _ PQ . PR . 

 PS . TT' ~ PS . PT' 

 NO. 1075, VOL. 42] 



Thus, the lines R'T' and RT are parallel ; so that, in order to 

 construct the ellipse, it is necessary to divide the two lines PT 

 and PQ by a series of parallel lines, meeting them in points T 

 and R, which, being projected from the centres 

 B and C respectively, will determine, by means 

 of the points of intersection of corresponding 

 rays, any number of points on the required 

 conic. 



Such is Newton's method of constructing a 

 conic five points on which are given. We will 

 nowproceed to prove the intimate connection 

 existing between it and the more modern 

 method illustrated in Fig. i, the discovery of 

 which has been sometimes attributed to Pascal. 

 It will be observed that when, as in Fig. 2, 

 the lines PT and PR are drawn parallel to the 

 adjacent sides AB and AC, the rays R'T' and 

 RT are parallel (5) ; and that, therefore, the 

 centre of perspectivity of the punctuated lines 

 PT and PR lies at an infinite distance. When, 

 however, the lines PT and PR are shifted into 

 the positions PT' and PR' (Fig. 3), being then 

 no longer parallel to the adjacent sides branch- 

 ing from A, the points DEO on the ellipse are 

 projected from B upon line PT' in d-^e-^o-^, and 

 from centre C upon the line PR' in d-^e^o^. 

 Now, in order to demonstrate the method 

 given at the beginning of this paper, it is 

 necessary and sufficient to show that the lines 

 joining d-^ and r//, e^ and e^, o^ and 0-1^, all 

 meet in one and the same point, S ; or, in 

 other words, that the punctuated lines PT' and PR' are in 

 perspective. 



Thus, if the transversal o^df^ be drawn parallel to PT, we 

 have — 



<hi(t_ — '^i)^ . <hej_ _ ^iP . 

 oe dB ' o^dQ eP 



and, if the transversal o^S be drawn parallel to PR', we have — 



Oi€ 



o,e. 



e.P 



wherefore, multiplying together the right and left hand members 

 of the last three equations, we obtain — 



(6) 



(7) 



(8) 



