164 
PROFESSOR M. J. M. HILL ON THE LOCUS OF SINGULAR POINTS 
Hence it touches the surface <f> ( x , y, z) + 2 - 7 = 0. 
Hence <f> (x, y, z) + - 2 - 7 - = 0 is an envelope. 
This accounts for the factor <f> (x, y, z) + - 2 - 7 - in the discriminant. 
(D). Examination of the term D 2 f l /DxDa 1 for this example. 
The equation Df/Da = 0 is, in this case, 
<P { x , V, 2 ) H bP {x, y, z) — aj = 0. 
Hence a 1 satisfies 
<P ( x = y> z) + 3 [> (x ; y , z) — aff = 0, 
and dafdx is determined by 
0(f) 
dx 
+ 6 [xp (x, y, z) — 
= 0 . 
Hence, at a point on the locus of binodal lines, i.e., where <f>(x, y, z) — 0, 
xp (x, y, z) — oq, it follows that da 1 /dx is infinite. 
Calculating D f/Dx D a, it follows that it is 
cd) r I r \ n 04r 
- Sx ->i»(x,y,z)-a]- d -. 
Hence, \}f l jDxDa l is equal to the value of — dcp/dx at the point on a locus ol 
binodal lines. Hence it is finite. 
(E). Examination of the values of f 2 
In this case a x , a. z are the roots of 
im 
/ 
<P {£, y, C) + 3 [xp (£ y, £) — aj = 0. 
Therefore, 
a i — 'Pity, 0 — v 7 {— £)}> 
% = ’A (£ y> C) + a/I — s <P (£ y, £)} ■ 
Hence 
f(x, y, z, a 2 ) = [<f> (£ 7], £)] 2 + <j> (£ y, £) [xp (£ y, £) — af\ + [xp (£ y, £) — o 2 ] 3 
= [<£ (£ ^5 £)] 2 ~ I (£ 0 v 7 ! — i <£ (£ *?> 01 
1 >ytey,,, a ,) = 6 w(fj ^ 0 - a,] = 6v /{- i </> (f,,,{)}. 
(£ ^ 0 
3/2 
— £ <P (£ ^ £) 
1 2 
<M£ ??> 0 + — £)}■ 
v 7 -r- 6 
0 T /a + 
1 
10V 
Hence 
