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508] IN PARTICULAR THOSE OF A QUADRIC SURFACE. 
and from these equations we deduce 
x _ <tt + 1 iy ar — 1 z r — a 
\fa t + a \/b r + a ’ \Jc t + cr' 
We have thence 
n T 
Vtt = ( t2 “ 1 ) da + - !) (-*-)» 
^ = (t 2 4-1) do* 4- (o- 2 + 1) dr O), 
~j~ c ~ —2rd(r+ 2adr (4-), 
where denom. = (t 4- cr) 2 : regarding 0-, t as the parameters in place of p, q, these show 
the values of the first differential coefficients a, a'; b, b'; c, c'. We deduce 
= — 2i Vic (<tt + 1) -4, B = — 2i'dca(aT — 1) 4-, G = — 2i \/ba (t — a) 4-, 
where denom. = (r + cr) 3 . We have, moreover, 
d 2 x 
\Jcl 
= — 2 (t 2 — 1) da 2 + 2 (err + 1) 2derdr — 2 (cr 2 — 1) dr 2 
2d 2 v 
— 2 (t 2 + 1) da 2 + 2 (err — 1) 2dadr — 2 (a 2 4- 1) dr 2 
4r da 2 + 2 ( t — cr) 2dadr 4- 
4crdT 2 
(+). 
(-). 
(-), 
where denom. = (t + <r) 3 : giving a, a', a"; /3, /3', /3"; 7, 7', 7". We deduce as the 
numerators of E' (= Aa + Bfi 4- Gy) and G' (= Aa" 4- Bf3" 4- Gy"), 
4i*Jabc {(ctt + 1) (t 2 — 1) — (ctt — 1) (t 2 + 1) — 2t (t — cr)}, = 0, 
and 
4i Vabc {(o-t + 1) (o- 2 — 1) — (or — 1) (a 2 + 1) + 2<r (t - <r){, = 0; 
that is, E’ — 0 and G' — 0; or the differential equation of the chief lines is dadr = 0, 
which is right. The value of F'(-Ad + B/3'+ Gy) is hardly required, but it is readily 
found to 
= 41 Vabc {— (ctt + l) 2 4- (ar — l) 2 — (t — cr)] 2 4- (r 4- cr) 6 , 
or since the term in { } is = — (t 4- a) 2 , we have 
— 4i V a&c 
F = . 
(t 4- o-) 4 
24. The values of E, F, G (ds 2 = Eda 2 4- 2Fdadr 4- Gdr 2 ) are 
E = a(r 2 — l) 2 — b (t 2 4- 1) 2 4- c. 4t 2 (4-), 
F = a (t 2 - 1) (cr 2 - 1) - b (t 2 + 1) (cr 2 4-1) - c. 4tct (-4), 
£ = a (a 2 - l) 2 - b (a 2 + l) 2 +c. 4cr 2 (-), 
where denom. = (r 4- cr) 4 . 
22—2