173 
508] IN PARTICULAR THOSE OF A QUADRIC SURFACE, 
and we ought therefore to have identically 
(p- q) ip %)) = T*j* + 2.4ArdT + Idr'- 4- (t + a-)‘; 
that is 
(p-qf = TS + (t + <t) 4 , 
p 2 — q 2 = A VTS (t + o-) 4 ; 
or, what is the same thing, 
(P - q y = TS (t + o-y, 
p + q = A -f- (t -j- o-) 3 , 
which are easily verified. 
26. In fact, the equation 
x 2 if z' 1 
f~ j 1 = 1 
Cl+ U Q + U G +11 
gives for u a quadric equation, the roots of which are u = p, u — q\ that is, we have 
p + q= a? + y 2 + z 2 — a— b — c, 
pq = — (b + c) x 2 — (c + a) y 2 — (a + b)z 2 +bc+ ca +ab; 
and substituting herein for x 2 , y 2 , z 2 their values in terms of a, r, we find 
p + q = a (a 2 — 1) (t 2 — 1) — b (a 2 + 1) (t 2 + 1) — 4cto- -t- (t 4- a-) 2 , 
pq = be (ar + l) 3 — ca((TT — l) 2 + ab (a — t) 2 -t-(t+o-) 4 , 
the first of which is, in fact, p + q = A -4- (r + a) 2 . And from the two equations, forming 
the combination (p + q) 2 — 4pq, we at once obtain the other equation 
(p - q) 2 = ST 4- (t + a) 4 . 
27. The most ready way of obtaining the relations between the differentials of 
p, q, a, t, is from the foregoing expressions of p + q, pq. Writing for a moment 
p +q=A -j- (cr -f- t ) 2 , pq = B + 4 t) 2 , we have p 2 (a + t) 2 - Ap + B= 0, q 2 {a+ r) 2 —Aq+B = 0; 
viz. the first of these equations is 
p 2 (a + t) 2 — p [a (a 2 — 1) (t 2 — 1) — b (cr 2 + 1)(t 2 + 1) — 4ccrr] 
+ be (ar + l) 2 — ca (err — l) 2 + ab (cr — r) 2 = 0, 
which is quadric in p, a, t. The negative discriminants in regard to these variables 
respectively are ST, 4PT, 4PS respectively, and we have thus the equation 
and the like equation for dq. 
dp 
VP 
+ 2 
/ da dr 
W2 + VT 
= 0; 
28. In the first integral of the geodesic lines, introducing a, t instead of p, q, 
the equation becomes 
V 
(da dr 
.p+e) Ws + vt 
r)V 
dir dr\ _ , 
.¡¡Zd)\j$~~ ’