Distance function of the Ellipse in $\mathbb{R}^n$ Let $\epsilon(a,b)$ denote the Ellipse in $\mathbb{R}^2$ i.e. the set $\lbrace (x,y): \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\rbrace $. 
Is there an explicit way to compute $f_{a,b}(x)=\operatorname{dist}(\epsilon(a,b),x)$ for points $x$ inside the ellipse boundaries? 
For the circle ($a=b=1$) one for example has $f(x)=r - \Vert x \Vert$. 
P.s. I want to show that $f_{a,b}(x)=\operatorname{dist}(\epsilon(a,b),x)$ has a unique maximum point and it is equal to $0$. Given this statement is true, maybe there is an easier way to see it than computing the function explicitly. 
 A: The ellipse can be parametrized as
$$x=a\cos\phi\\
y=b\sin\phi$$
Then, the distance from a point in $\mathbb{R}^2$, $p=(x_0,y_0)$, to any point to the ellipse is
$$d^2(p,\epsilon)=(x_0-a\cos\phi)^2+(y_0-a\sin\phi)^2$$
The distance can be minimize (or maximize) by taking the derivative with respect to the angle, $\phi$, finding
$$(a^2+b^2)\sin\phi\cos\phi=ax_0\sin\phi-by_0\cos\phi$$
This equation will provide the maximum and minimum distance from the point to the boundary of the ellipse
note
This equation can be transformed into a fourth order equation, by using the changes of variable $\cos\phi=\frac{1-t^2}{1+t^2}$ and $\sin\phi=\frac{2t}{1+t^2}$.
A: 
I want to show that $f_{a,b}(x, y) = \operatorname{dist}(\epsilon(a, b), (x, y))$ has a unique maximum point and it is equal to $(0, 0)$.

(I've taken the liberty of writing points as ordered pairs in your question statement.)
Without loss of generality, say $0 < b < a$. For each $x_{0}$ with $-a < x_{0} < a$, there exists a unique $0 < y \leq b$ such that $(x_{0}, \pm y)$ lie on $\epsilon(a, b)$. These points are separated by a distance of $2y \leq 2b$, with equality if and only if $x_{0} = 0$.
If $(x_{0}, y_{0})$ is inside the ellipse, then $|y_{0} - y| + |y_{0} + y| = 2y \leq 2b$. It follows at once that
$$
f_{a, b}(x_{0}, y_{0}) = \min(|y_{0} - y|, |y_{0} + y|) \leq b,
$$
with equality if and only if $(x_{0}, y_{0}) = (0, 0)$.
