200 PROFESSOR M. J. M. HILL ON THE LOCUS OF SINGULAR POINTS 
Now, transforming the equation by means of the substitutions x = a + X, 
y = b -|- Y, 2 = cr — 6 2 4- Z, it becomes 
«X 3 + /3Y 3 - 3 n (Z - «X + 6Y) 2 = 0. 
Hence, the new origin is a unode. 
Hence the locus of unodes is 2 = x 2 — y 2 , i.e., £ — 0. 
(C.) The Envelope Locus is 
{9 ay 8£ + 4n (/3z 3 - a?/ 3 )} 2 + 64 a/ Sn 2 a;V = 0 .... (94). 
On examining the manner in which the discriminant was formed it can be seen 
that the factor corresponding to the envelope focus is obtained from (86), (87), and 
3£Z + aX - yY = 0. 
Hence it is the result of eliminating X, Y, Z from 
£Z -j- xX. — yY — 
aX 2 + AnxtJZL = 
/3Y 2 - 4n«/£Z 2 = 
2£Z' 
0 
0 
(95). 
To prove that (94) is an envelope locus, it is necessary to show that if <f> be the 
left-hand side of (94), and f = 0 he the equation of the system of surfaces, then it is 
possible to find values of x, y, z which satisfy at the same time all the equations— 
f — 0, <j> = 0 
iy /fy _ Df /ty _ Iff /fo b 
D xj dx D yj oy D zj 3 z j 
(96). 
Changing the independent variables from x, y , 2 to x, y, £, then, if § denote partial 
differentiation when x, y, £ are independent variables, equations (96) are equivalent to 
i.e 
/=0, cf) — 0 
/H>_ _ §f /%4> _ 3 / /8j> 
bx/ bx ~ by I by ~ bz/ bz ’ 
3 a (x — a ) 2 — 6?i (2 x — a) [% -I- (x — a)x — (y —b) y ] 
24 n/3x 2 [9«/9£ + 4?t (/9a; s — a?/ 3 )] + ld2n 2 a^x 2 y i 
3/9 (?/ — b) 2 + (2?/ •—&)[£ + (x ~ a) x ~ (y —b) y\ 
— 24/ia?/ 3 [9a/9£ -)- 4w (/9x 3 — c«/ 3 )] + 19’2n 2 u/3x?y 2 
— 6w [^ + a; (cc — a) — y (y — 6)] 
