1891-92.] Dr Muir on a Problom of Elimination. 
29 
( a 2 - 62 ) 2 {(^2 + ^ 2 ) _ (^2 + /, 2 ) J 4 = Q , 
as it should be. 
Secondly, for the case ^vhere the tracing point is a focus of the 
ellipse, that is, where 'p^ = a^ - IP- and g' = 0, we find 
0 = 0, X = 
r = 4(«2 - 62)2 _ ^ _ 4 ^ 2 ^ 2 ^ 
0 = 0, o-= 262; 
= 1 6a2(cfc2 _ 
p = 2cP, 
and the equation takes the form 
1 6(a2 - 62)2/aj2 ^ _j_ y2^^2y2 _ + h^{x^ + y2) | = 0, 
i.e.^ (^2 _|_ y2\^i^2y2 ^ ^ 4 ^ _ 4 ^ 2 _^- 2^2 . 
which is readily seen to be correct, because the original equations 
then become 
2px cos 6 + 62 - aj 2 ^ Q 'I 
2py sin 6 + 62 - y2 = 0 J" 
and thus give 
a;2-262 + ^ + j,2_252 + -^ = 4»2 = 4a2-4J2 
i.e., 
ipp + y2)(^2y2 q. ^4J ^ \:OpX^y‘^. 
Thirdly, there is the very interesting case where the tracing point 
is an extremity of an axis ; say, where = 0, g = 6. In this case 
we find 
O = a'i, p = a2 + 262, 
r = (a2 - 262)2, 
d> = 3a4-l6a262 + 1664, 
0 = a6, 
^ = Zap - - ZaPb^ + 1 66^, 
O' = u2^ 
X = 
A = + 4a262, 
A = 3a8-4a662-12a46^; 
and the equation takes at the outset the rather awkward form 
