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)

332 
ON THE CENTRO-SURFACE OF AN ELLIPSOID. 
[520 
The Nodal Curve. Art. Nos. 29 to 60. 
29. If two different points on the ellipsoid correspond to the same point on the 
centro-surface, this will be a point on the Nodal Curve: the conditions for this if 
(P v)> (p> Vi) are the parameters for the two points on tho ellipsoid, are obviously 
(a 2 4- |) 3 (a 2 + rj) = (a 2 4- p) 3 (a 2 4- Vi), 
(b 2 + p) 3 (b 2 +v) = + P ) 3 (b 2 + Vi)> 
(c 2 4- |) 3 (c 2 + v) = (c 2 + p) 3 (c 2 + 77^; 
these equations in effect determine 77 as a function of p so that the equations 
— ¡3ya 2 x 2 = (a 2 4- p) 3 (a 2 + 77), &c. 
then determine the coordinates (x, y, z) of a point on the Nodal Curve in terms of 
the single parameter p 
The relation between £ and 77 would be obtained by eliminating p, 77 : from the 
foregoing equation: but it is easier to eliminate 77 and thus obtaining between 
P and £ a relation in virtue of which ¿5 may be regarded as a known function of 
£; 77 and 77j can then be expressed in terms of p ¿5, so that each of these quantities 
will be in effect a known function of £( x ). 
30. The relation between p p is in the first instance given in the form 
a 2 [(a 2 4-1) 3 — (a 2 4- p) 3 ], (a 2 + £) 3 , 
b 2 [(6 2 + P 3 - (b 2 + p) 3 ], (V + Z) 3 , 
c 2 [(c 2 4- f) 3 - (c 2 4-10 3 ], (c 2 4- p) 3 , 
(a 2 + P) 3 
(6 2 + P) 3 
= 0. 
(c 2 4- P) 3 
Throwing out a factor (f—p) 2 , this becomes 
2 [a? [3a 4 4- 3a 2 (f 4- p) 4- p 4- f p 4- p 2 } 
X(6»-C ! ).(l, 1, !]№ + f)(# + (,), (i> s + f,)(c 5 + ?)>] = <>, 
where the left-hand side is a symmetrical function of p p vanishing for f = p, and 
therefore divisible by (£ — p) 2 ; it is also divisible by A, = (Z) 2 — c 2 ) (c 2 — a 2 ) (a 2 — b 2 ) (= a/3y). 
To work this out, write f 4- P = y>, £p = the equation may be written 
2 {(6 2 — c 2 ) a 2 
3 a 4 
4- 3a 2 /) 
4-_/) 2 - # 
3 Z> 4 c 4 
4- 36 2 c 2 (6 2 4- c 2 )/) 
4- (Z> 4 4- c 4 ) (p 2 - 5») 
4- 6 2 c 2 (/) 2 4- 8</) 
4- 3 (b 2 + c 2 )pq 
4- 3 q 2 
where the left-hand side divides by A (p 2 — 4q). 
1 = 0, 
1 This was my first method of solution; and I have thought the results quite interesting enough to 
retain them—but it will appear in the sequel that I have succeeded in expressing £, 17, £ 1} -q x , in terms of a 
single parameter <r.
	        
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