AND LINES IN THE INTERSECTIONS OF A SYSTEM OF SURFACES 
265 
f(x, y, z, a, b) = 0, 
0, 
Bf 
Da 
w 
D b 
= 0. 
Let £ + X, y + Y, £ + Z be a neighbouring point on the same three surfaces, so 
that the values of a , b are the same. 
Therefore 
/+ [flX + M Y + raz + 1 {[£ f]X» + • • • } 
+ * (K ^ A X 8 + = 0 (183), 
[a] + [£ a]X + [^]Y + [U]Z + HKf ) «]S 3 + ...}+... = 0 (184), 
Da + K fl X + [v, fi]Y + l£,fi]Z + t {[£ 6 0] X 2 + o (185). 
Hence because f — 0, [a] = 0, [/3] = 0, the terms of lowest order in X, Y, Z in 
(183), (184), (185) are of the first degree in each case. Hence there is one solution 
X = 0, Y = 0, Z = 0. Hence there is one intersection at this point. 
(B.) Next consider the case where the contact is stationary. 
The equation of the tangent plane to the envelope locus is 
(x-a[a + (Y-^)M + (z-ora = o .... ase). 
But also from (28) and (29) # by means of (76) the equation of the tangent plane 
can also be shown to be 
[“> ft] {(X — f) [a, f] + (Y — 1 1) [a, 5)] + (Z — Cl [a, {]} 
- [«, a] {(X - f) [ft f] + (Y - v ) [/3, ,] + (Z - {) 1/3, £]} = 0 (187). 
Hence 
ili {[«. ft] [«. f] - [«, «] [A Cl = ^ {[a, (3] [a, ,] - [«, a] [ft ,]} 
= [jj {[*> ft] [“> 0 - [*.»] [A {]}• 
Hence the lowest terms in (183), (184), (185) are not independent of each other; 
and if the three fractions last written be each = y, it is possible by multiplying (183) 
* Equations (28) and (29) are satisfied at any point of an envelope locus. Equations (16), (17), 
(18) are not. 
MDCCCXCII. —A. 2 M 
