476 DR G. PLARR ON THE DETERMINATION OF 



Hence (as &j<f>~\j = r, &c.) 



u 2 y 2 — 2u 2 vy = b 2 [w 2 — a 2 — b 2 ] 



u 2 z 2 — 2u 2 wz = b 2 [v 2 —a 2 — b 2 "] . 

 But 



w 2 —a 2 —b 2 = l 1 —u 2 —v 2 -a 2 — b 2 ; 

 as 



l l = "La 2 = a 2 + 2b 2 

 therefore 



w 2 — a 2 — b 2 = b 2 — u 2 — v 2 



v 2 — a 2 —b 2 = b 2 — u 2 — w 2 . 



Putting every term into the first members of the preceding equations, and ordaining, we 



get the quadrics, 



b 2 v 2 - 2u 2 yv + [u 2 y 2 + b%u 2 - b 2 )] = 



b 2 w 2 - 2u 2 zw + [u 2 z 2 + b 2 (u 2 - b 2 )] = . 



Hence in resolving as to v, iv, 



b*v =u 2 y + Y | 



b 2 w = u 2 z+Z J ' 



where 



Y2 = u y _ &[ U Y + &2( u 2 _ &2 )] t 



Z 2 = u% 2 - b 2 [u 2 y 2 + b\v? - &)] , 

 or also 



Y 2 = {u 2 -b 2 ){u 2 y 2 -¥), 



Z 2 ={v?-b 2 )(v?z 2 -¥) . 

 Eliminating v, w from both of their values we get 



2y{u 2 y + Y) = ~[t + \ + 2n 2 ^\ 



20 (u 2 z + Z) = ^ [t + L - 2u 2 R] . 



4u 4 R 2 = R' 2 ,2u 2 R = R, 

 R' 2 = (t + L ) 2 - 8mu 6 s(t + L + Lj ) 



4w2Yi/-6 2 R'=6 2 (i + L )-4uY . . . (I.) 



\u 2 Zz + 6 2 R' = b\t + L ) - 4u 4 2 . (II.) 



L o = -(S^+m 2 ), 

 -Si0i = 6 2 [Si0 - H + i\a 2 + b 2 )] 



Putting 

 namely, 

 we get 



We have for L ( 

 and as 



m 



= b 2 [u 2 -(a 2 + b 2 )] 

 = ( - a 2 V)2 ( - ^\ = ¥ + 2a 2 b 2 ., 



