x² y² 2aex(x² + y²) b²(x² + y²)²
——— + ——— + ——————————————— - ———————————————— = 0.
a² b² a² a²
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