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)

[10 
48 
ON THE THEORY OF ALGEBRAIC CURVES. 
the series in { } being continued only to x~ mn+1 , because the terms after this point 
produce in the whole product only terms involving negative powers of x. It is for 
the same reason that the series in ( ), in the equations (3) and (4), are only continued 
to the terms involving x~ m+1 , x~ n+1 respectively. 
The first side of the equation (6) is of the order mn, in x, as it ought to be. 
But it is easy to see, from the form of the expression, in what case the order of the 
first side reduces itself to a number less than run. Thus, if n be not greater than m, 
and the following equations be satisfied, 
A = a, A (1) = a (1) ...A (r-1) = r n 
B = /3, Bu = ... B< s -1) = ^(s-i) } 
K = k, R (1) = ac (1 > ... K (v ~ l) = v >> u, 
the degree of the equation (6) is evidently mn - r — s ... - v, or the curves U = 0, V = 0 
intersect in this number only of points. If mn — r — s... - v = 0, the curves ¡7=0 and 
V=0 do not intersect at all, and if mn — r—s — v be negative, = — w suppose, the 
equation (6) is satisfied identically; or the functions ¡7, V have a common factor, the 
number <y expressing the degree of this factor in x, y. 
Supposing the function V given arbitrarily, it may be required to determine U, so 
that the curves ¡7=0, V = 0 intersect in a number mn—k points. This may in general 
be done, and done in a variety of ways, for any value of k from unity to (ra + 3). 
1 shall not discuss the question generally at present, nor examine into the meaning 
of the quantity mn — \m (m + 3) { = \m (2n — m — 3)} becoming negative, but confine 
myself to the simple case of U and V, both of them functions of the second order. 
It is required, then, to find the equations of all those curves of the second order 
which intersect a given curve of the second oi’der in a number of points less than four. 
Assume in general 
and reducing, we obtain 
V = (y — Ax — A') {y — Bx — B') + K. 
Similarly assume 
U = {y — ax — a!) (y — /3x — /3') + k.
	        
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