214 
PROFESSOR M. J. M. FULL ON THE LOCUS OF SINGULAR POINTS 
(C.) r lhe Locus a.8~z -fi 4y 3 = 0 is an Ordinary Envelope. 
This can be proved by finding the tangent planes parallel to 2 = 0 . 
Hence it is necessary to satisfy at the same time 
a (x — af ~b 3/3 (y — b)' : ~f 3y (x — a) z + S' 
d = 0, 
« {x — af ~b y 2 
= 0, 
y — b 
= 0, 
Hence 
3y (x — a) ~b 2 £2 
f 0 
therefore 
11 
0 - 
a. (x — a) 3 ~b 3y (x — a) z - b Sz 2 = 0, 
a (x — - a) 2 4- yz =0. 
Hence 
2y (x — a) z -fi S 2 3 = 0. 
(124). 
The solution 2 = 0 of the last equation makes x — a, and does not satisfy (124). 
Hence it is necessary to take 
2 y (x — a) + 8z = 0 . 
This gives 
Hence, when 
scS 3 2 -fi 4y 3 = 0 . 
X = a -b 2y 3 /aS, y = b, Z = —= 4y 3 /a S', 
the tangent plane is parallel to the plane 2 = 0 . It touches all the surfaces of the 
system. 
Hence 
% = — 4y 3 /ctS- 
is an ordinary envelope, 
Art. 15.— To prove that if the. Edge of the Biplanar Node always touch the Biplanar 
Node Locus, then A contains B 3 as a factor . 
(A.) It will be shown that when the condition (76) holds in the case of a biplanar 
node locus, then the edge of the biplanar node always touches the biplanar node locus. 
The equation of the biplanes is given by (33). 
Now, if the left-hand side of (33) break up into two linear factors, then the two 
planes, whose equations are given by equating the two linear factors to zero, will 
intersect in the straight line whose equations are given by any two of the three 
equations;— 
