186 
PROFESSOR M. J. M. HILL ON THE LOCUS OF SINGULAR POINTS 
It may be noticed that z — 0 touches each of the surfaces at one point only, 
x = ct, y — b. This result may be compared with the next example. 
Example 3. — Ordinary Envelope. 
Let the surfaces be 
z + (x — a) (x~ -f y~ — b) — 0. 
(A.) The Discriminant. 
A = z. 
(B.) The Envelope Locus is z — 0. 
Every point on z = 0 is the point of contact of one only of the surfaces. Hence 
z occurs as a factor once only in the discriminant. 
It may be noticed that z = 0 touches each of the surfaces at two points, viz., 
x = a, y = ± y/(b — a 2 ). 
Art. 10.— To prove that if C = 0 be the equation of the Conic Node Locus, A contains 
C 2 as a factor in general. 
Let f, y, £ be a point on the conic node locus, then equations ( 11 ) —(14) are satisfied. 
Hence, by (50), the substitutions x — y = rj, z = £, make A = 0 ; and, by (53), 
they also make 3A/3a: = 0. 
By symmetry they also make 3A/3 y = 0 , 3A /dz = 0 . 
Hence, by Art. 1 , Preliminary Theorem B, A must contain C 2 as a factor. 
Example 4.— Locus of Conic Nodes. 
Let the surfaces be 
a (x — a ) 3 + fi (y — b) z + 6 m (x — a) (y — b) + y - 2 = 0 , 
where a, ft, y, m are fixed constants, a, b are the arbitrary parameters. 
(A.) The Discriminant. 
To eliminate a and b between f — 0 , D//Da = 0 , D//D5 = 0 , is, in this case, the 
same as eliminating x — a, y — b between 
f =z 0 —--— = 0 — = 0 • 
J ’ D O -a) 5 D {y - h) 
i.e. , making the equation homogeneous by putting 
x — a — X/Z, y — b = Y/Z, 
it is necessary to find the discriminant of 
<*X 3 + /3Y 3 + 6 mXYZ 4- yz 2 Z 3 = 0 . 
