Parallelepiped $P$ circumscribed around ellipsoid $S$: length of main diagonal of $P$ depends only on $S$, formula. Let $S$ be an ellipsoid in $\mathbb{R}^n$ (as in my previous question here), and let $P$ be a parallelepiped such that $S$ is inscribed in $P$ (in particular, each face of $P$ is tangent to $S$). However, we do not require the edges of $P$ to be parallel to the axes of $S$; in particular, $S$ does not determine $P$ unique.


*

*What is the easiest way to see that the length of the main diagonal of $P$ depends only on $S$?

*What is a formula for this length in terms of the coefficients $a_1, \dots, a_n$ introduced by my aforementioned previous question here?

 A: $
\newcommand{\n}[1]{\overrightarrow{\boldsymbol{n}}_{#1}}
\newcommand{\v}{\overrightarrow{\boldsymbol{v}}}
\newcommand{\Q}{\overrightarrow{Q}}
\newcommand{\T}[1]{T_{#1}}
\newcommand{\x}{x_0}%{\chi}
\newcommand{\y}{y_0}%{\gamma}
$
If I understand your question correctly, then the statement 

$\,\ldots\,$ the length of the main diagonal of $\,P\,$ depends only on $\,S\,$

is not correct. 
I assume that by main diagonal you mean the largest of the space diagonals. 
Then the length will depend not only on the equation for $\,S,\,$ but also on the choice of points of tangency.

General equation in $\,2D\,$
For demonstration purposes, let us restrict ourselves to $\,2D\,$ case: 

Given ellipse $\,S\,$  inscribed into parallelogram $\,P\,$ we want to find the length of its main diagonal.

In order to construct parallelogram $\,P\,$ we need to 


*

*Choose two different points on ellipse $\,S\,$ 

*Write out tangent lines for this points and for their opposites


*

*this will give us equations for the lines on which edges of $\,P\,$ lie


*Find  points of intersections for these lines   


*

*this will give us coordinates of vertices of $\,P\,$


*Compute the maximum distance between vertices,     


*

*this will give us the length of main diagonal of  $\,P\,$



It is possible to show that the maximum distance will depend on the choice of  tangency points, although it requires some tedious calculations. 
Instead, let me write out general equations for edges of $\,P\,$ and use them in a counterexample.

General equations
Let $\,S\,$ be an ellipse with semi axes $a$ and $b$ inscribed into  a parallelogram $\,P$:
$$
S = \left\lbrace \big( x,y \big) \in \mathbb R^2 \; \Big\vert \; \ 
\frac{x^2}{a^2} + \frac{y^2}{b^2}  = 1 \right\rbrace
\quad \text{ with the gradient } \quad
\nabla S = \begin{bmatrix} \frac{2}{a^2} x \\ \frac{2}{b^2} y \end{bmatrix}
$$
Let us denote $\,\T Q\,$ the line tangent to  $\,S\,$ at a point $\,Q = \big(\,\x, \y\,\big)\,$
\begin{align}
\T Q  
& = \left\lbrace \,  \v  \in \mathbb R^2 \,\Big\vert \;\,
   \n Q  \cdot  \left( \v - \Q \right)  = \overrightarrow 0\, \right\rbrace
\\ 
& = \left\lbrace\begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb R^2 \; \Big\vert \; \ 
\begin{bmatrix} \frac{2}{a^2} \,\x \\ \frac{2}{b^2} \,\y \end{bmatrix}
\cdot 
\begin{bmatrix} x-\x \\ y-\y \end{bmatrix}
=
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\right\rbrace
\\ 
& = \left\lbrace \big( x,y \big)  \; \Big\vert \; \ 
\frac{2\x}{a^2} \big(x - \x \big) + \frac{2\y}{b^2} \big(y - \y \big)  = 0 \right\rbrace
\end{align}
\begin{align}
\text{where} \qquad \qquad 
\v   = \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb R^2, 
\qquad
\n Q = \Delta S \,\Big\vert_{Q} 
= \begin{bmatrix} \frac{1}{a^2} \,\x \\ \frac{1}{b^2} \,\y \end{bmatrix},
\qquad 
\Q = \overrightarrow{OQ} = \begin{bmatrix} \x \\ \y \end{bmatrix}
. \qquad \qquad  \qquad \qquad \qquad\qquad
\end{align}

Counterexample
Assume that semi axis of $\,S\,$ are $\,a=2\,$ and $\,b=1,\,$ then 
$%\begin{align}
\ S = \left\lbrace \big( x,y \big) \in \mathbb R^2 \; \Big\vert \; \ 
\frac{x^2}{2^2} + \frac{y^2}{1^2}  = 1 \right\rbrace
.%\end{align}
$
Case 1:
Denote tangent points as 
$\,A=\big(\,-2,0\,\big),\,$
$\,B=\big(\,0, 1\,\big),\,$
$\,C=\big(\, 2,0\,\big),\,$
$\,D=\big(\,0,-1\,\big).\,$
Then the vertices of $\,P_1\,$ are 
$\,I=\big(\,-2,-1\,\big),\,$
$\,J=\big(\,-2, 1\,\big),\,$
$\,K=\big(\, 2, 1\,\big),\,$
$\,L=\big(\, 2,-1\,\big),\,$
Then the length of main diagonal is equal to $\,2\sqrt{5}.\,$ 

Case 2:
Denote tangent points as
$$\,M = \left(\, -\sqrt{2},  \dfrac{\sqrt{2}}{2}\,\right),\quad
\,N = \left(\, -\sqrt{2}, -\dfrac{\sqrt{2}}{2}\,\right),\quad 
\,O = \left(\,  \sqrt{2}, -\dfrac{\sqrt{2}}{2}\,\right),\quad 
\,P = \left(\,  \sqrt{2},  \dfrac{\sqrt{2}}{2}\,\right).$$    
Then the vertices of $\,P_2\,$ are
$$\,E = \left(\, -2\sqrt{2}, 0 \,\right),\quad
\,F = \left(\, 0, \sqrt{2} \,\right),\quad
\,G = \left(\, 2\sqrt{2}, 0 \,\right),\quad
\,H = \left(\, 0, -\sqrt{2} \,\right).$$    
Then the length of main diagonal is equal to $\,4\sqrt{2}.\,$ 

Clearly in the second case the main diagonal is larger.
