The inverse of a conic from any point not on the curve is a nodal bicircular quartic. This is shown by inverting the general equation of the conic
ax² + 2hxy + by² + 2gx + 2fy + c = 0 ;
| x | y | ||
| by substituting for x and y | ——— | and | ——— , |
| x² + y² | x² + y² |
we get the equation
ax² + 2hxy + by² + 2(gx + fy)(x² + y²) + c(x² + y²)² = 0.
The origin is evidently a double point on the curve, and is a crunode, acnode, or cusp according as the conic is a hyperbola, ellipse, or parabola. The factors of the terms of the fourth degree, viz: (x + iy) (x + iy) (x - iy) (x - iy), show that the two imaginary circular points at infinity are double points on the quartic, which is thus trinodal. Hence this nodal, bicircular quartic can be projected into the most general form of the trinodal quartic. Trinodal quartics are unicursal.
If the conic which we invert be a parabola, the quartic has two nodes and one cusp. If the conic be inverted from a focus, the quartic has the two circular points at infinity for cusps. This is best shown analytically as follows: let the equation of the conic, origin being at the focus, be written
| x² | y² | 2aex | b² | ||||
| —— | + | —— | + | —— | - | — | = 0. |
| a² | b² | a² | a² |
Inverting this we have
| x² | y² | 2aex(x² + y²) | b²(x² + y²)² | ||||
| —— | + | —— | + | —————— | - | ————— | = 0. |
| a² | b² | a² | a² |