1898-99,] Lord M‘Laren on Glissette Elimination Problem. 381 
the solution evidently is of the form, 
0 = a 2 (ll 2 — a 2 ft 2 ) + ft 2 (m 2 — a 2 ft 2 ) + y 2 (7 2 — a 2 ft 2 ) + 2aft(laft — mn) + 2ay(lllaft - 111) | 
+ 2fty{naft — Im) j 
and, by putting for 7, m and n their equivalents, we have 
0 = a 4 (A - y 2 ) 2 + /3 4 (A - X 2 ) 2 + y 4 (A - C) 2 + 2a 2 /3 2 y(A - C) + 2a 2 /3 2 y(A - X 2 ) 
+ 2a 2 /3 2 y(A — f 2 ) 
- a 2 ft 2 . a 2 - a 2 ft 2 , ft 2 - a 2 ft 2 y 2 - 2a 2 /3 2 (A - y 2 )( A - £ 2 ) - 2a 2 y 2 (A - y 2 )(A - O') 
- 2/3 2 y 2 (A — z 2 )(A - C) 
where each vertically-placed pair of terms corresponds to a double 
term in the preceding formula. 
The 1st and 2nd terms of the 1st line with the 4th term of the 
2nd line form a square factor, which I shall put last in order. 
Then arranging by powers of (A - C), y, and aft the equation is 
(A - C ) 2 y 4 -2(A-C)y 2 {a 2 (A-y 2 )+ft 2 (A-X 2 )} +2(A-C)a 2 /3 2 y 
- a 2 ft\a 2 + ft 2 + y 2 ) + 2a 2 /3 2 y(2A - x 2 - if) + { a 2 ( A - y 2 ) - ft\ A - X 2 ) } 2 = 0 
But observing that a = 2 roc, ft = 2ry , the highest powers of the 
square term disappear, and the term reduces to A 2 (a 2 — ft 2 ) 2 ; 
whence, 
A o = 0 = (A - C) V - 2(A - C)/V(A . -if) + £2(A -**)} + 2(A- C) a 2 /J 2 y ) 
- a 2 /3 2 (a 2 + y8 2 + y 2 ) + 2a 2 ^ 2 y(2 A - x 2 - f) + A 2 (a 2 - /3 2 ) 2 j 
Comparing the complete and the abbreviated forms of the original 
equations, (1) and (2), we have A == a 2 —r 2 ; C = b 2 ; a = 2rx; ft = 
2ry • y = A + C -x 2 -y 2 = a 2 -r 2 + b 2 - x 2 -y\ 
If the tracing-point be at the focus, we have 
a 2 — r 2 -b 2 = A- C = 0; y = 2A -x 2 -y 2 . 
The terms in the first line of the expression, A 0 = 0, which are all 
multiplied by (A - C), vanish ; and the eliminant becomes, after 
substituting for y, 
0 = - a 2 /! 2 {a 2 + ft 2 + (2 A - a 2 - y 2 ) 2 } + 2a 2 ft\2A - x 2 - y 2 ) 2 + A 2 (a 2 - ft 2 ) 2 
= a 2 ft 2 {{2A -X 2 - y 2 ) 2 - a 2 - ft 2 } + A 2 (a 2 - ft 2 ) 2 
= a 2 /3 2 {4A 2 + (x 2 + y 2 ) 2 - 4A(x 2 + y 2 ) -a 2 - ft 2 } + A 2 (a 2 - ft 2 ) 2 
= a 2 ft 2 {(x 2 + iff - 4A(x 2 + y 2 ) - a 2 - ft 2 } + A 2 (a 2 + ft 2 ) 2 
( by combining the first and last terms of the preceding expression), 
= 1 6 r 4 x 2 y 2 {(x 2 + iff - 4(A + r 2 ){x 2 + if) } + 1 6r 4 A?(x 2 + y 2 ) 2 
