ON THE CURVES WHICH SATISFY GIVEN CONDITIONS. 
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[406 
of the sextic surface leads to the following values (agreeing with those obtained from 
the formulae by writing therein m = n = 2, a = 6), viz. 
(1 : :) =6, = 2m + n, 
(1, 1 .'.) = 4, = 2m 2 + 2 mn + f n 2 — 2m — \n — fa, 
(2.-.) =6, = a, 
(3 :) =4, = — 4m — 3n + 3a. 
I remark that the section by an arbitrary plane is a sextic curve having 6 cusps and 
4 nodes; it is therefore a iinicurscil sextic; this suggests the theorem that the sextic 
surface is also unicursal, viz. that the coordinates are expressible rationally in terms of 
two parameters ; I have found that this is in fact the case. In doing this there is 
no loss of generality in supposing that a = b = c = 1; and assuming that this is so, 
and putting also — 1 + h = k, 1 +h — k 1} and therefore 2h = k + k lf we have 
A — — 3 kki, 
B = — kk\w — (k + ki) z, 
G = — x 2 — y 2 — z 2 + (k + h) (xy — zw), 
D — 3z (2xy — zw). 
The equation of the sextic surface being, as before, 
A 2 D 2 + 4>AC 3 + 4 B 3 D - 3 B 2 G 2 - 6 A BCD = 0, 
I say that this equation is satisfied on writing therein 
~2 
+ y = Y"~3fc(i“ ^ sin 
- y = \/\ (1 + k a) cos <p, 
= 1, 
w = ^2a — co8?(f) + ^2a — sin 2 <p, 
where (a, </>) are arbitrary. In fact these values give 
%A = — kk\ cos 2 cp — kk x sin 2 <£, 
B = — k (2a&! + 1) cos 2 cf) — k\ (2ak +1) sin 2 <p, 
C = — k ( ak\ + 2) cos 2 <p — k\a ( ak + 2) sin 2 (p, 
£D = — ka 2 cos 2 (f> — k x a 2 sin 2 (p, 
whence, <w being arbitrary, we have 
±(A, B, C, D$o>, l) 3 
= — [A: cos 2 <p (k\co + 1) + k\ sin 2 (p (kco + 1)] (a) + a) 2 ,