AND LINES IN THE INTERSECTIONS OF A SYSTEM OF SURFACES. 
189 
Example 5 .—Locus of Biplanar Nodes. 
Let the surfaces be 
a (x — of + a (y — 6) 3 + 6 m (x — a — z) (y — b — z) = 0 , 
where a, m are fixed constants; a, b the arbitrary parameters. 
(A.) The Discriminant. 
r 
With the notation of the last article, the discriminant is the same as that of the 
equation 
aX 3 + aY 3 + 6m (X - zZ) (Y -zZ)Z = 0. 
Therefore 
S = 2m 2 aV — m 4 . 
T = 36m 2 a 4 z 4 — 64m 3 a 3 z 3 -f* 24m 4 a 2 z 2 — 8 m 6 . 
A = 16m 4 a 3 z 3 ( 9az + 4m) (3 a 2 ^ 2 — 6 maz + 4m 2 ) 2 . 
(B.) The Biplanar Node Locus is z = 0. 
Transforming the equation by the substitutions x = a + X, y = b + Y, z — Z, it 
becomes 
«X 3 + aY 3 + 6m (X - Z) (Y - Z) = 0. 
Hence the new origin is a biplanar node on the surface. Hence z = 0 is the locus 
of biplanar nodes. 
(C.) (i.) The Ordinary Envelope is 9 az + 4m = 0 . 
(ii.) The Envelope such that every point on it is the point of contact of two 
non-consecutive Surfaces is 3a 2 z 2 — 6 maz + 4m 2 = 0 . 
To prove these statements it is necessary to find the tangent planes parallel to the 
plane z = 0. 
Hence it is necessary to have 
x.e., 
j = 0, D//Dx = 0, D f/Oy = 0, Df/Dz =f= 0, 
a (x — a) 3 + a (y — 6) 3 + 6m (x — a 
a (x — a) 2 + 2m (y — b — z) 
a (y — 6) 2 -f 2 m (x — a — z) 
) (y - b - z) = ° . 
■ (69). 
= 0 . 
■ (70). 
= 0 . 
• (71). 
0 . 
■ (72). 
x — a — z -\- y — b — z 
