Tschirnhausen cubic



In algebraic geometry, the Tschirnhausen cubic, or Tschirnhaus' cubic is a plane curve defined, in its left-opening form, by the polar equation
 * $$r = a\sec^3 \left(\frac{\theta}{3}\right)$$

where $a = 1$ is the secant function.

History
The curve was studied by von Tschirnhaus, de L'Hôpital, and Catalan. It was given the name Tschirnhausen cubic in a 1900 paper by Raymond Clare Archibald, though it is sometimes known as de L'Hôpital's cubic or the trisectrix of Catalan.

Other equations
Put $$t=\tan(\theta/3)$$. Then applying triple-angle formulas gives
 * $$x=a\cos \theta \sec^3 \frac{\theta}{3} = a \left(\cos^3 \frac{\theta}{3} - 3 \cos \frac{\theta}{3} \sin^2 \frac{\theta}{3} \right) \sec^3 \frac{\theta}{3}= a\left(1 - 3 \tan^2 \frac{\theta}{3}\right)$$
 * $$= a(1 - 3t^2) $$
 * $$y=a\sin \theta \sec^3 \frac{\theta}{3} = a \left(3 \cos^2 \frac{\theta}{3}\sin \frac{\theta}{3} - \sin^3 \frac{\theta}{3} \right) \sec^3 \frac{\theta}{3}= a \left(3 \tan \frac{\theta}{3} - \tan^3 \frac{\theta}{3} \right) $$
 * $$= at(3-t^2)$$

giving a parametric form for the curve. The parameter t can be eliminated easily giving the Cartesian equation
 * $$27ay^2 = (a-x)(8a+x)^2$$.

If the curve is translated horizontally by 8a and the signs of the variables are changed, the equations of the resulting right-opening curve are
 * $$x = 3a(3-t^2)$$
 * $$y = at(3-t^2)$$

and in Cartesian coordinates
 * $$x^3=9a \left(x^2-3y^2 \right)$$.

This gives the alternative polar form
 * $$r=9a \left(\sec \theta - 3\sec \theta \tan^2 \theta \right)$$.

Generalization
The Tschirnhausen cubic is a Sinusoidal spiral with n = &minus;1/3.