1898 - 99 .] Lord M‘Laren on Glissette Elimination Problem. 385 
In forming the final determinant we eliminate the factor 
{(A - C) cos </> sin </> + B(sin 2 </> - cos 2 </>)} in one column, thus, 
(") 
(b) 
y (c) 
If; ® 
This, it will be observed, is a bordered symmetrical determinant 
of the 4th order. 
By cancelling the terms multiplied by B (the coefficient of cos </> 
sin <j> in the original equations) we get the determinant already 
found for an axial tracing-point, as we ought to do. *■ 
The developed solution is 
[cos </>] 
afi — 2By 
(A-C)y 
fi(A-X 2 )-Ba 
[sin <£] 
[ 1 ] 
( (A - C) cos cf> sin </> ) 
( + B (sin 2 </> - cos 2 </>) j" 
(A - C)y fi(A - x 2 ) - Ba 
afi + 2By a(A - y 2 ) 4 - Bfi 
a(A -y 2 ) + Bfi a fi 
fi y 
0 (A r ) 
a 2 (Aa — ay 2 + B /3) 2 
+ fi 2 (Afi - fix 2 - Ba) 2 
+ y(A — C) 2 
- 2afi(A fi - fix 2 - Ba)(Aa - ay 2 + Bfi) 
- 2ay 2 (A - C)(Aa - ay 2 + B fi) 
+ 2/3y 2 (A - Q>)(Afi - fix 2 - Ba) 
— ct?fi(afi + 2By) 
-afi\afi-2By) 
— a 2 /3 2 y 2 + 4B 2 y 4 
+ 2a 2 fi 2 (A - C)y 
+ 2ay(a fi + 2By)(A/3 — fix 2 — Ba) 
+ 2fiy(afi - 2By)(Aa - ay 2 + Bfi) 
We can form a complete square out of the terms in the 
first column if we substitute for the 4th term, 
(2afi-iafi)(Afi - fix 2 — Ba)(Aa - ay 2 + Bfi). 
Thence, 
{ a(Aa - ay 2 + Bfi) + fi(Afi - fix 2 - Ba) - (A - C )y 2 } 2 - 4a fi(A fi-fix 2 - Ba)(Aa - ay 2 + ByS) 
+ (a/3 + 2By){2ay(A/3 — fix 2 - Ba)-a *fi} + (afi — 2By) { 2/3y( Aa — ay 2 + B/3) — a/3 3 } 
+ 4B 2 y 4 — a 2 /3 2 y 2 + 2( A — ■ C)a 2 fi 2 y = 0 (A,) 
The expression found for the glissette having its tracing-point 
on the axis was not combined precisely in this way, because the 
reduction to the 6th degree for the tracing-point at focus was more 
easily effected by keeping the eliminant in its original form. 
It would be easy to expand A 1 by writing out the coefficients 
at full length, but such an expansion would serve no useful pur- 
pose. In fact, the bordered symmetrical determinant, after sub- 
