THE CURVE ON ONE OF THE COORDINATE PLANES. 



487 



+ b 2 



4 1 



g(«2 + <V- a l)— 3O2 + «s) 



(because of a 2 = ^, a' 2 = § ). Hence 



— ^-i = « 2 a 1 3 (a 1 — a 2 X a i — a 3 ) 



17, Ct, 



6 2 

 + g[(a 2 + %) + Sa^aaaffg - 2)] . 



If now we had a x = a 2 — —a 3 , then (as they are all three of the same absolute value), the 

 coefficient of a 2 would vanish, but that of b 2 would be negative 



= ^(-3^-2) 



and Mj could not vanish. 



If a x = — a. 2 = —a 3 , then the coefficient of b 2 would become 



^2a 1 [-l + 3a 1 2 -2] = ^| ) 



4 

 whereas the coefficient of a 2 would become + - , so that M x would vanish only for a 2 — b'\ 



J v3 



There exists therefore the only solution 



j_ l l 

 a 1 = a 2 = a 3 =-\ — -=, or = -=, 



which annuls M 1( namely, the direction for a of the median line of the octant i, j, k (or 

 its opposite direction, which does not alter the circumstances), 



The values of y and z now become equal, that is 



— - n\ 



1 =»&-»)]. 



As 



This gives 



l 1 -n = a?+2b 2 -(a 2 - b 2 ) = 3& 2 

 &V3 



2 2 , 2 6&4 



and 



t = ti 2 r 2 = 2b*. 



We have thus a system of values of t, u 2 , namely 2& 4 and \l x respectively, which solve 



the equation (V"). It is, however, a certainty that this point y = z = — ^=. does not 



v n 



belong to the outward branch of the limiting curve, but that it forms a singular point of 

 a branch (or branches) contained in the inside of the limiting curve as defined by \\ie = 0. 



VOL. XXXV. PART II. (NO. 13). 4 L 



