Find the shortest distance between $\begin{align}
&\frac{x_1^2}{a_1^2}+…+\frac{x_n^2}{a_n^2}=1\\
&\frac{x_1^2}{a_1^2}+…+\frac{x_n^2}{a_n^2}=2
\end{align}$.
My thinking is when $n=2$, we have two ellipses where the axes of the second are the same as the first scaled by a factor of $\sqrt{2}$. WLOG, assume $|a_1|<|a_2|$. Then the shortest distance would be $|a_1|(\sqrt{2}-1)$. For the ellipsoid case, I would say the shortest distance is $|a_{\text{min}}|(\sqrt{2}-1)$. I don't know how to prove this formally. I thought about using Lagrange multipliers but had difficulties even setting it up. Any suggestions?
Best Answer
In general, the minimum and the maximum distances (with respect to the standard norm $\|\bullet\|$ on $\mathbb{R}^n$) between two $n$-dimensional ellipsoids given by $$A:=\Biggl\{\left(x_1,x_2,\ldots,x_n\right)\in\mathbb{R}^n\,\Big|\,\sum_{i=1}^n\,\frac{x_i^2}{a_i^2}=1\Biggr\}$$ and $$B:=\Biggl\{\left(x_1,x_2,\ldots,x_n\right)\in\mathbb{R}^n\,\Big|\,\sum_{i=1}^n\,\frac{x_i^2}{b_i^2}=1\Biggr\}\,,$$ where $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ are positive real numbers such that $a_i\geq b_i$ for every $i=1,2,\ldots,n$, are $$\min\big\{a_1-b_1,a_2-b_2,\ldots,a_n-b_n\big\}$$ and $$\max\big\{a_1+b_1,a_2+b_2,\ldots,a_n+b_n\big\}\,,$$ respectively. (Of course, the minimum distance is $0$ if there exist $j,k\in\{1,2,\ldots,n\}$ such that $a_j\geq b_j$ and $a_k<b_k$, whilst the maximum distance is still given by the same formula above.) To show this, one only needs to observe that, if $\textbf{u}\in A$ and $\textbf{v}\in B$ are such that $\|\textbf{u}-\textbf{v}\|$ is optimal, then $\textbf{u}-\textbf{v}$ is parallel to the normal vector of $A$ at $\textbf{u}$ as well as the normal vector of $B$ at $\textbf{v}$.