92 
Proceedings of Royal Society of Edinburgh, [sess. 
X and V are of the 2nd degree in the £i?-and-y coordinates of the 
curve, a, /?, A, and /x are of the 1st degree in these coordinates. 
B, C are functions of constants only. 
These values being compared with the determinant, it is seen 
that the eliminant is of the 8th degree in x and y. The subjoined 
tables give respectively the highest and lowest indices of the 
coordinates, x and y, for each term of the determinant : — 
0 0 11 
..11 
112 . 
11.2 
2 2 3 3 
= 0 : 
fo 
0 ^ 1 
1 
0 
0 1 
. 1 
1 0 
1 . 
0 1 
1 0 
1 0 
. 1 
0 1 
1 0 
The first table shows the eliminant to be of the 8th degree. 
The absolute term is given by the zero indices of the second 
table. It is included in the term (B^ + of which the absolute 
part is 
{{V^ - + cff + {(P + 5 ^ . 
This quantity may vanish by either factor becoming =0, when 
the curve passes through the point of intersection of the guides. 
For the case of + the curve has its principal axis 
in the 1st and 3rd quadrants. The other case apparently repre- 
sents a curve whose axis of symmetry is in the 2nd and 4th 
quadrants. 
A solution has been given by Dr Muir in the current volume 
of the Proceedings^ vol. xix. p. 25, which was obtained by a 
different process. His eliminant as first obtained is of the 10th 
degree; and the extraneous factor being found, the equation is 
then reduced to the 8th degree, and expanded in terms of x 
and y. 
It is proposed to verify the two solutions by comparing the con- 
stituent terms of the 8th degree in my solution with the corresponding 
terms as given by Dr Muir. The first step is to multiply out the 
factors of the eliminant above found, (A), p. 90, arranging the 
terms according to powers of the non-homogeneous elements y and v. 
The resulting expression is 
