AND LINES IN THE INTERSECTIONS OF A SYSTEM OF SURFACES. 247 
Equating the third ratio of (£) to this, and substituting for 2 from (e), and putting 
a/x = £, b/y = 77, the result is 
m (m — c) £ + m (m -f- c) 77 + 2 ( g 3 — m 3 ) = 0. (77). 
In like manner, from the first and third ratios of (£), 
m 3 (c 3 — g 2 ) yf + 4m (m + c) /q + 4 (gd — m 3 /) 
-f- £ {— m 3 (c 3 + /) 17 + 2 m/ (m — c)} = 0 . . . (0). 
Substituting for £ from equation (77), this reduces to 
mcr )' 2 + rj (g 2 — cm) = 0. 
Hence 77 = 0, 77 = 1 — ( g 2 jcm ). 
Substituting in equation (77) the corresponding values of £ are 
, _ 2 (m 3 - f) 
m (to — c) ’ 
I = 1 + (//cm). 
It remains to prove that one of the two systems of solutions will satisfy the equa¬ 
tion obtained from the second and third ratios of (£). 
This equation is 
m 3 (c 3 — /) £ 3 + 4 m (m — c) + 4 (g^ — m 2 g 2 ) 
+ 77 [ — m 3 (c 3 + /) £ + 2m/ (m + c)] = 0. 
Substituting for 77 from equation (77), this reduces to 
mc£ 3 — £ (cm + /) = 0. 
Therefore 
£ = 0, £ = 1 + (//cm). 
Hence the solutions 
x = acm/(cm + /), 
y = hem I {cm — p' 2 ), 
« = 2a6c 3 /m/{(c 3 m 3 — p' 4 ) (c 2 — /)}, 
satisfy all the equations. 
Hence the surface (e) is an envelope. 
(D.) It will now be verified that the values of b given by (181), and the equations 
obtained by changing £ into 77 and £, become equal in this case. 
