Recall that the polar of a set $A\subset\mathbb{R}^n$ is the following set: $$A^{\circ} = \lbrace c \in \mathbb{R}^{n} |\; \langle c,x \rangle \leq 1 \quad \forall x \in A \rbrace$$ where $\langle \cdot, \cdot \rangle$ is the usal scalar product. The polar of a centred ball of radius $r$ is a centred ball of radius $1/r$. In particular the unit centred ball (for the euclidean norm) is the polar of itself.
Now, I am wondering what is the polar of a non centred ball (containing the origin)?
My guess is that it is an ellipsoid but I am not sure how to prove that. Does someone knows the answer and how to prove it, or has a good reference?
In favor of my conjecture , I did some construction with GeoGebra. I constructed a regular polygon $P$ with many sides to approximate the ball. I constructed the polar of $P$. The vertices of the dual of $P$ all lie on an ellipse, whatever the position of $P$ is, as long as it contains the origin. This is not a proof, but it makes me convinced that, at least in dimension 2, the conjecture holds.
My idea to prove the conjecture is to see the non centred ball $B\subset\mathbb{R}^n$ as a cross section of a higher dimensional cone $C\subset\mathbb{R}^{n+1}$ and the polar $B^\circ$ as a cross section of the dual cone $C^\circ$. More precisely, we set $\hat{B}:=B\times\{1\}\subset\mathbb{R}^{n+1}$ and $C$ the cone generated by $\hat{B}$. Then we can show that $\check{B}:=B\times\{-1\}\subset\mathbb{R}^{n+1}$ is the cross section of $C^\circ$ with the hyperplane $\mathbb{R}^n\times\{-1\}$.
I remember from school that an ellipse can be viewed as intersection of a cone with an hyperplane. That why the construction of the previous cones seems appropriate. Nevertheless, it is not completely clear to me what happens here. In particular I don't see well if the cone build from this non centred ball is of the same kind as a cone which is constructed from a centred ball.