Radial distance for superellipsoid

Question

Consider a point $[x_0\,\,y_0\,\,z_0]'\in\mathbb{R}^3$ and define the inside-outside function \begin{equation*}F(x_0,y_0,z_0)\triangleq \left[\left(\frac{x_0}{a_1}\right)^{2/\varepsilon_2}+\left(\frac{y_0}{a_2}\right)^{2/\varepsilon_2}\right]^{\varepsilon_2/\varepsilon_1}+\left(\frac{z_0}{a_3}\right)^{2/\varepsilon_1}\end{equation*} where $a_1$,$a_2$, $a_3$, $\varepsilon_1$, $\varepsilon_2$ are nonnegative parameters. Now define $\beta$ such as the scaling factor that satisfies \begin{equation*}F(\beta x_0, \beta y_0, \beta z_0)=1\end{equation*} I'm not able to proove that, according to the previous definitions, follows \begin{equation}F(x_0,y_0,z_0)=\beta^{-2/\varepsilon_1}\end{equation} I would like to have some hints regarding how to proove this result.

Answer 1

Note that:

\begin{align*} 1=F(\beta x_0,\beta y_0,\beta z_0)&=\left[\left(\frac{\beta x_0}{a_1}\right)^{2/\epsilon_2}+\left(\frac{\beta y_0}{a_2}\right)^{2/\epsilon_2}\right]^{\epsilon_2/\epsilon_1}+\left(\frac{\beta z_0}{a_3}\right)^{2/\epsilon_1}\\ &=\left[\beta^{2/\epsilon_2}\left(\frac{x_0}{a_1}\right)^{2/\epsilon_2}+\beta^{2/\epsilon_2}\left(\frac{y_0}{a_2}\right)^{2/\epsilon_2}\right]^{\epsilon_2/\epsilon_1}+\beta^{2/\epsilon_1}\left(\frac{z_0}{a_3}\right)^{2/\epsilon_1}\\ &=\beta^{2/\epsilon_1}\left[\left(\frac{x_0}{a_1}\right)^{2/\epsilon_2}+\left(\frac{y_0}{a_2}\right)^{2/\epsilon_2}\right]^{\epsilon_2/\epsilon_1}+\beta^{2/\epsilon_1}\left(\frac{z_0}{a_3}\right)^{2/\epsilon_1}\\ &=\beta^{2/\epsilon_1}F(x_0,y_0,z_0) \end{align*} and hence $$F(x_0,y_0,z_0)=\beta^{-2/\epsilon_1}$$