How can one calculate the pedal curve with respect to the origin of the Folium of Descartes,
$x^3+y^3-3axy=0$.
Note that an implicit curve is desired, not its parametrization. I already know how to calculate a pedal curve, however the pedal to this curve has escaped my grasp to compute. I know the answer should be
$a^2 (3 a^2 x^2 y^2 + 6 x y (x^2 + y^2)^2 + 4 a (x^5 + x^3 y^2 + x^2 y^3 + y^5)) = (x^2 + y^2)^4$
But how do you compute this?
I got as far as
$\frac{x_0^2-ay_0}{ax_0y_0} = \frac{\cos(\theta)}{r}$
and
$\frac{y_0^2-ax_0}{ay_0x_0} = \frac{\sin(\theta)}{r}$
Where $x_0$ and $y_0$ are points that are in the curve, meaning
$x_0^3+y_0^3 -3ax_0y_0 = 0$
And $r$ and $\theta$ are the polar form of the pedal curve.