222 
PROFESSOR M. J. M. IIILL ON THE LOCHS OP SINGULAR POINTS 
a (x — 
a ) 3 
+ 
P(y — 
b) s -f 3 [c (x — 
ci) + e(y - 
-b) + gzf - 
- hh 2 = 0 
(138), 
a(x — 
af 
+ 
2 c [c (x 
— a) + e (y — 
b) + gz] 
= 0 
(139), 
P(y - 
■bf 
+ 
2 e [c {x 
— a) 4 - e (y — 
h ) + gz] 
= 0 
(140), 
3 g [C (x 
- a) + e(y - 
■ 6) + gz] - 
- h 2 z 
(1«). 
or putting x 
— a 
= 
X/Z ,y 
I 
C - 1 
II 
Isi 
c+- 
rD 
s equations 
(138)— (140) become the 
same as 
(133), (134), (135). 
Hence the result of the elimination will be the same as in the previous case. It is 
only necessary to show that (141) is satisfied. 
Multiplying (139) by (x — a), (140) by (y — b) and subtracting from (138), it 
follows that 
[c ( x — a) -f e (y — b) + gz] [c (x — a) + e (y — b) + 3 gz] — hh 2 — 0 . 
Therefore 
[c (x — «) + e (y — b) + gzf + 2gz [c (x — a) -f e (y — b) + gz] — hh 3 = 0 . 
Therefore 
c (x — a) + e (y — b) + gz = — gz ±2 \/((/ 3 + ^ 2 ). 
Hence (141) is not satisfied unless z — 0 . 
Now 2 = 0 makes c (x — a) + e (y — b) + gz — 0 . 
Therefore x = a by (139) and y = b by (140). 
This solution corresponds to the binode locus. 
It may therefore be excluded. 
Hence the factor of the discriminant under discussion corresponds to an envelope 
locus, touching all the surfaces ; it consists of four planes parallel to 2 = 0 , whose 
equations are independent of the arbitrary parameters. 
Section IV. (Arts. 16-25). —Consideration of cases reserved from the previous 
section. The degree of f (x, y, z, a , b) in a, b is now the second, and the 
equations Dfi/Da = 0, Df/Db = 0 are indeterminate equations for the 
PARAMETERS AT POINTS ON THE LOCUS OF ULTIMATE INTERSECTIONS. 
It was supposed in the previous section that the degree of/ ( x, y, z, a, b) in a, b was 
higher than the second ; for if the degree were the second, and the analytical condition 
satisfied which expresses that at a point on the locus of ultimate intersections, two 
systems of values of the parameters, which satisfy D/’/Do = 0 , Df/Db = 0 , become 
equal, then this analytical condition requires to be specially interpreted. 
For now Df/Dct = 0 , Df/Db = 0 are two simple equations in «, b. Hence they are 
either satisfied by one value of a and one of b, or else are indeterminate. But since 
the condition 
