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.
 

If you look at the proof that you linked, the proof is using an Affine transformation.
Now you can also see that the centroid of the triangles are mapped to each other in the affine transformation, and so do the centers of the ellipses.
Thus the ellipse you need has the same center as your triangle! This property uniquely defines the ellipse with the maximum area.
Now consider the points $A = (-1,-4/3), B = (2, -4/3)$ and $C= (-1, 8/3)$. This is a 3-4-5 triangle whose center is origin which was gotten by starting with $(0,0), (3,0)$ and $(0,4)$ and translating so that the centroid is the origin.
Now the equation of an ellipse whose center is the origin is given by
$Px^2 + Qxy + Ry^2 = 1$.
Thus we must have that
(1) $P + 4Q/3 + 16R/9 = 1$ 
(2) $4P - 8Q/3 + 16R/9 = 1$ 
(3) $P -8Q/3 + 64R/9 = 1$
Solving these (see footnote) gives us the equation of the ellipse as
$x^{2}/3 + xy/4 + 3y^{2}/16 = 1$
In order to verify this, the area of $Px^2 + Qxy + Ry^2 = 1$ is given by $\displaystyle \frac{2\pi}{\sqrt{4PR - Q^2}}$ which comes out to $\displaystyle \frac{8\pi}{\sqrt 3}$

You can ignore the below if you like. This is just manually solving the equations
(1) $P + 4Q/3 + 16R/9 = 1$ 
(2) $4P - 8Q/3 + 16R/9 = 1$ 
(3) $P -8Q/3 + 64R/9 = 1$
Subtracting (1) and (2) gives $3P = 4Q$.
Subtracting (2) and (3) gives $3P = 48R/9$
Thus $Q = 3P/4$ and $R = 9P/16$
Thus using (3) we have that
$P - (8/3)*(3P/4) + 64/9 * (9P/16) = 1$
i.e
$P - 2P + 4P = 1$ i.e $P = 1/3$.
