Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., sadlerian professor of pure mathematics in the University of Cambridge (Vol. 1)

382 
ON GEOMETRICAL RECIPROCITY. 
[61 
m = BG -F 2 , 3$ = CA-G 2 , (B=AB-H 2 , 
§ = GH-AF, (& = HF-BG, ffi = FG-CH, 
© = ^U 2 + 33 m 2 + (Bn 2 + 2jpmn. + 2CBrnZ + 2 |i=^m, 
□ = (&2 4- + <3m) (72/ - fiz) 
+ (P^ + 33 m + $ n ) i az - y x ) 
+ (<8xl + Jpm + (Bn) (Ax - ay) ; 
suppose U = 0 represents the equation of the polar conic, U — P 2 = 0 that of the pole 
conic. The two tangents drawn to the polar conic are represented by UU 0 — IT 2 = 0, and 
by determining Jc in such a way that UU 0 — W 2 — k (TJ — P 2 ) may divide into factors 
the equation 
UU 0 - W 2 — k(TJ — P 2 ) = 0, 
represents the lines passing through the points of intersection of the tangents with the pole 
conic. Thus if Jc~TJq, the equation reduces itself to U 0 P 2 —W 2 = 0, or W = ± V(Go) P, 
the equation of two straight lines each of which passes through the point of inter 
section of the lines P—0, W = 0, (that is, of the line of contact of the conics, and the 
ordinary polar of the point with respect to the polar conic); these are in fact the 
lines A 1 B 2 , A 2 B 1 intersecting in the line of contact. The remaining value of k is not 
easily determined, but by a somewhat tedious process I have found it to be 
= K(TJ 0 -P 0 2 ) + (®-K). 
In fact, substituting the value, it may be shown that 
□ 2 + K(PP 0 - W) 2 = K(U 0 -Pf)(U-P 2 ) + (0 -K)(TJU 0 - W 2 ), 
which is an equation of the required form. To verify this, we have, by a simple 
reduction, 
®(W 0 - W 2 )-[J 2 = K(UP 0 2 -2WPP 0 +U 0 P 2 )-, 
or, writing for shortness 
7y- fiz = g, az — yx = y, Ax - ay = £ 
{m 2 + 33m 2 + (Bn 2 + 2 jfmn + 2&nl + Tffilm) (gtf* + 33r + (B? + 2jfy£ + 2ffi(£ + Hfagy) 
— [@Ut: + 3$my + (Bn£ + jp (ni1 + ra£) + (5r (Ig + ng) + (mg + ly)] 2 
= K {A (mg — ny) 2 + B (ng — lg) 2 + G (ly — mg) 2 
+ 2F (ng — Ig) (ly — mg) + 2G (ly — mg) (mg — ny) -1- 2H (mg — ny) (ng — 1%)}, 
which is easily seen to be identically true.
	        
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