Minimal Ellipse Circumscribing A Right Triangle Find the equation of the ellipse circumscribing a right triangle whose lengths of it's sides are $3,4,5$ and such that its area is the minimum possible one.
You may chose the origin and orientation of the $x,y$ axes as you want.
Motivation: It can be proved [Problem of the Week, Problem No. 8 (Fall 2008 Series), Department of Mathematics, Purdue University] that the area of this ellipse is $8\pi /\sqrt{3}$, without the need of using its equation, but I am also interested in finding it.

Edit: picture from this answer.
 

Given the proof to which you linked, as Rahul Narain said in a comment, there is an affine transformation that maps the unit circle and an inscribed equilateral triangle to the desired ellipse for a particular 3-4-5 right triangle.  Represent this affine transformation by
$$\begin{pmatrix}x'\\\\y'\\\\1\end{pmatrix}=\begin{pmatrix}a_{1,1} & a_{1,2} & b_1\\\\a_{2,1} & a_{2,2} & b_2\\\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\\\y\\\\1\end{pmatrix}.$$
Let the 3-4-5 right triangle have vertices (0,0), (0,3), and (4,0), and let the equilateral triangle have vertices (1,0), $(-\frac{1}{2},\frac{\sqrt{3}}{2})$, and $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$.  Mapping these vertices in the order listed and solving for the parameters of the transformation gives the matrix
$$\begin{pmatrix}-\frac{4}{3} & -\frac{4}{\sqrt{3}} & \frac{4}{3}\\\\-1 & \sqrt{3} & 1\\\\0 & 0 & 1\end{pmatrix}.$$
The unit circle has parametric representation $x=\cos t$, $y=\sin t$.  Applying the affine transformation to this unit circle gives the parametric representation $x=\frac{4}{3}(1-\cos t-\sqrt{3}\sin t)$, $y=1-\cos t+\sqrt{3}\sin t$ for the desired ellipse.

edit: Looking at Moron's answer, his choice of x and y coordinates are swapped relative to mine, and as he said, his right triangle is translated to have its centroid at the origin, but accounting for these differences, my parametric equations satisfy his equation.  That is, based on his equation, an equation in x and y alone for the ellipse with parametric equations as I specified is $9x^2+12xy+16y^2-36x-48y=0$.

edit 2: I don't know too much of the detailed mechanics of the different representations of affine transformations.  The form that I used above is equivalent to
$$\begin{pmatrix}x'\\\\y'\end{pmatrix}=\begin{pmatrix}a_{1,1} & a_{1,2}\\\\a_{2,1} & a_{2,2}\end{pmatrix}\begin{pmatrix}x\\\\y\end{pmatrix}+\begin{pmatrix}b_1\\\\b_2\end{pmatrix}$$
and the 1s and 0s in the third row in the earlier version are just the glue that allows an affine transformation to be represented with a single matrix multiplication.  My solution could have just as easily used this representation of the affine transformation.
