196 PROFESSOR M. J. M. HILL OH THE LOCHS OF SINGULAR POINTS 
Example 7 .—Locus of Uniplancir Nodes. 
Let the surfaces be 
a (x —■ of + fi (y — bf — 3 n (z — ax 1 -f- by) 2 = 0, 
where a, fi, n are fixed constants, and a, b are the arbitrary parameters. 
(A.) The Disci - iminont. 
Putting £ = z — x 2 -f- if, the equation is 
a(x — cif + fi(y — b) s — 3 n {£ + x (x — a) — y (y — b)}~ = 0. 
Hence the discriminant is the same as that of the equation 
aX 3 + fiY* - 3nZ (£Z + xX - yY) 2 = 0. 
Therefore 
S = rdafixy'Q 
l 1 = 9n 2 a 2 /P£ 4 + 4n 3 a/3£ 3 (fix* — ay 3 ). 
Therefore 
A = ?d'a 3 /3 2 £ G [{9a/3£ + 4 n (fix* — ay 3 )} 2 -f* 64?ra/3x 3 y 3 ]. 
In order to show the way in which the factor £ G arises in the discriminant, the dis¬ 
criminant will now he calculated. 
It is known to be the result of eliminating X, Y, Z from Df/D% = 0, D//DY = 0, 
D//DZ =• 0, i.e., from 
aX 3 — 2nxZ (£Z + xX —■ yY) = 0 , 
■ (86), 
y8Y 2 +2nyZ(£Z+xX-yY) = 0 ...... 
. (87), 
(£Z + xX - yY) (3£Z + xX - yY) = 0 . . . 
. (88). 
From (86) and (87) 
Y = ± x v / (-|). 
. (89). 
Substituting in (86) 
aX 3 - 
- 2 nxXZ j x T y \/( - “) } - 2w(Z 3 = 0. 
Put 
X — 
<* 
I 
^14 
II 
jfrv 
+ 
<< 
1 
II 
-p 
Then two of the values of X/Z are found from 
aX 2 — 2 nx^XZ — 2 »x£Z 2 = 0 
(90), 
