Now transform this equation so that the lines joining the origin to the circular points at infinity shall be the axes of reference. To do this

let x + iy = x₁ and x - iy = y₁;

x1 + y1 x1 - y1
∴ x = ———  and  y = ———
2 2i

Making these substitutions and reducing we have (dropping the subscripts),

(x² + 2xy + y²) (x² - 2xy + y²) 4aexy(x + y) b²x²y²
—————— - —————— - —————— - ————— = 0.
a² b² a² a²

Making this equation homogeneous by means of z, we have

(x² + 2xy + y²) (x² - 2xy + y²) 4aexy(x + y) b²x²y²
z²—————— - —————— - —————— - ————— = 0.
a² b² a² a²

which is the equation of the quartic referred to the triangle formed by the three nodes. We are now able to determine the nature of the node at the vertex (y, z). Factor x² out of all the terms which contain it; and arrange thus:

z² z² 4aexyz b²y²
x²— - — - ——— - ——
a² b² a² a²
yz² yz² 2aexy² z
+ 2x — + — - ———
a² b² a²