109 
576] “ON THE RAY-PLANES IN BIAXAL CRYSTALS.” 
The required condition is that the coefficient of u 2 shall vanish ; viz. we then have 
that is, 
— afiy sec 2 6 = aÇ~ + yS 2 
= (b — c) (a + b) 2 + (a — b)(b + c) 2 
= (a — c) {36 2 + b (a + c) — ac} 
= - /3 (4i> 2 + 07), 
ay sec 2 6 = 4b 2 + ay, 
agreeing with Mr Walton’s value. Giving sec 2 6 this value, and throwing out the 
factor u, the equation becomes 
u {2a 3 ay£ + 2y 3 ca8 + ay (ca — b 2 ) 2 — ary 2 (4b 2 + ay)\ 
+ y (a 2 a 3 + c 2 y 3 ) = 0; 
or, what is the same thing, 
ayu {2a (a + b) (6 — c) 2 + 2c(c + b) (b — a) 2 + (ca — b 2 ) 2 — (b — c)(a— b) 4b 2 — (b — c) 2 (b — a) 2 } 
+ y (cl 2 a? + c 2 7 s ) = 0, 
say 
ay Ku + (a 2 a s + c 2 7 3 ) y = 0, 
viz. this equation determines the remaining intersection of the curve by the line 
yz — ax—0] the point in question lies outside the triangle, that is, u : y should be 
negative; or a, 7, a 2 a 3 + c 2 y 3 being each positive, we should have K positive; we in fact 
find 
which is as it should be. 
K = 4 Z> 4 + b 2 (a 2 + c 2 — 6ac) + 4a 2 c 2 
= 4 (b 2 — ac) 2 + b 2 (a + c) 2 ,