Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 8)

94 ON A THEOREM RELATING TO EIGHT POINTS ON A CONIC. [500 
In fact this equation written with an indeterminate coefficient X, say, for shortness, 
thus 
67 [(12.45) (23.56)] = X12. [(23.56) (34.67)] = 0, 
is the general equation of the conic through the 4 points 12.67, 34.67, 12.45, and 
23.56; and by making the conic pass through 1 of the remaining 4 of the 8 points, 
I succeeded in finding the value \ = ^—so that the conic in question passes 
through 5 of the 8 points, and is therefore by the theorem the conic through the 
8 points. But as thus written down the equation contains the extraneous factor 2—6, 
as appears at once by the observation that the left-hand side on writing therein 
6 = 2 (a 6 = a2) becomes identically = 0 ; the value in fact is 
- (2 - 8) [x - (2 + 7) y + 27*] (23 - 34 + 45 - 52) [x - (1 + 2) y + 12*] 
+ (2 - 8) [x - (1 + 2) y + 12*] (23 - 34 + 45 - 52) [x - (2 + 7) y + 27*] 
which is = 0; there is consequently the factor 2—6 to be rejected, and throwing this out 
the equation assumes a symmetrical form in regard to the 8 symbols 1, 2, 3, 4, 5, 6, 7, 8. 
The coefficient of a? is very easily found to be 
= (2 - 6) (12 - 23 + 34 - 45 + 56 - 67 + 78 - 81), 
and similarly that of * 2 to be 
= (2 - 6) [123456 - 234567 + 345678 - 456781 + 567812 - 678123 + 781234 - 812345} : 
those of the other terms are somewhat more difficult to calculate; but the final result, 
throwing out the factor (2 — 6), and introducing an abbreviated notation 
212 = (12 — 23 + 34 — 45 + 56 - 67 + 78 - 81), 
and the like in other cases, is found to be 
« 2 . 212 
+ y 2 . [212 (4 + 5) (6 + 7) — \ 21256] 
+ * 2 . 2123456 
- yz . 216 (234 + 235 + 245 + 345) 
+ *«.[21234 +J 21256] 
— xy . 212(4 + 5 + 6 + 7) =0, 
where it is to be observed that 21256 consists of 4 distinct terms each twice repeated: 
^ 21256 consists therefore of these 4 terms; and in the coefficient of y 2 they destroy 
4 of the 32 terms of 212 (4 + 5) (6 + 7) so that the coefficient of y 2 contains 
32 — 4, =28 terms. In the coefficient of zx there is no destruction, and this contains 
therefore 12 + 4, =16 terms.
	        
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