NOTE ON THE CASSINIAN. 
265 
601] 
To verify that there are on the axis only two ordinary foci, we write in the 
equation x = az + iy, and determine a by the condition that the resulting equation for 
y (which equation, by reason that the circular point z= 0, x—iy, is a node, will be 
a quadric equation only) shall have two equal roots; the equation is in fact 
{(a - a) 2 z 2 + 2 (a-a) iyz} {(a + a) 2 z 2 - 2 (a + a) iyz} - c 2 z* = 0, 
viz. throwing out the factor z 2 , this is 
(a 2 — a 2 ) {(a — a) z + 2iy) {(a + a) z + 2iy\ - c*z 2 = 0, 
or, what is the same thing, it is 
(a 2 — a 2 ) {(az + 2iy) 2 — a 2 z 2 j — c 4 z 2 = 0, 
viz. it is 
(2iy + olz) 2 - (a 2 + 2 2 = 0. 
The condition in order that this may have equal roots is 
a 2 + —- = 0, that is, a 2 = a 2 — ; 
a 2 —a 2 a 2 
hence a has only the two values + 
viz. there are only two ordinary foci. 
C. IX. 
34