I'm looking for an equation of the pedal curve of a generalised superellipse given by: $$f(x,y)=\left|\frac{x}{a}\right|^m+\left|\frac{y}{b}\right|^n-1=0, \quad m,n>0,\qquad (1)$$ the pedal point is the origin $(0,0)$. I want the pedal curve equation to be expressed in polar coordinates $p = f(\theta)$. For simplicity it can be calculated only for the first quadrant of the coordinate system (absolute value can be omitted in the $f(x,y)$).
I know how to calculate it for $m=n$:
Following pedal curve definition, we start from finding tangent line to the super elipse in the point $\left(x_t,y_t\right)$: $$ \left(y - y_t\right)\left.\frac{\partial f}{\partial y}\right|_{x_t,\,y_t} + \left(x - x_t\right)\left.\frac{\partial f}{\partial x}\right|_{x_t,\,y_t} =0 , $$ $$\frac{n}{b^n}y_t^{n-1} y - \frac{n}{b^n}y_t^{n} + \frac{n}{a^n}x_t^{n-1}x - \frac{n}{a^n}x_t^{n} = 0 \qquad /\frac{1}{n},$$ $$\frac{1}{b^n}y_t^{n-1} y + \frac{1}{a^n}x_t^{n-1}x= \left(\frac{y_t}{b}\right)^n + \left(\frac{x_t}{a}\right)^n,$$ $$\frac{1}{b^n}y_t^{n-1} y + \frac{1}{a^n}x_t^{n-1}x= 1.\qquad (2)$$
Next, assume that:$$p = x\cos\theta+y\sin\theta,\qquad (3)$$ touches the tangent. By comparing $(2)$ and $(3)$ we can find: $$x_t = \left(\frac{a^n\cos\theta}{p}\right)^\frac{n}{n-1}, \quad y_t = \left(\frac{b^n\sin\theta}{p}\right)^\frac{n}{n-1},\qquad(4)$$ putting $(4)$ to $(1)$ we obtain desired equation of the pedal curve: $$p = \left(\left(a\cos\theta\right)^{\frac{n}{n-1}}+\left(b\sin\theta\right)^{\frac{n}{n-1}}\right)^{\frac{n-1}{n}}.$$ This is a typical schem of calculation, one can find in text books. Unfortunately, it is not so easy to use it for $m\neq n$, because I can't get the equvialent of the equations $(4)$ $(x_t = ...\,,y_t =...)$.
Does anyone know how to find this pedal curve?
Thanks in advance!
It seems unlikely that you will find a nice closed-form solution to this problem. However, the following may be of some help. First of all, we can express the generalized superellipse as follows:
$$x(t)=|\cos t|^{\frac{2}{m}}\cdot a\ \text{sgn}(\cos t)\\ y(t)=|\sin t|^{\frac{2}{n}}\cdot b\ \text{sgn}(\sin t)\\ z(t)=x(t)+iy(t)$$
Zwikker (see below) gives the pedal equation in the complex plane as follows:
$$z_p=\frac12\frac{z \dot z^*-z^* \dot z}{\dot z^*}$$
where the asterisk denotes the complex conjugate. I have tried this numerically and it seems to work, at least on the few simple cases I have tried.
Ref: Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.