Minimal Ellipse Circumscribing A Right Triangle

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 $P = 1/3$.

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.

Not much to contribute at this point except this enhanced picture:

(dashed lines: axes; gray lines: medians)

and the Mathematica notebook used to produce it.

(Thanks Isaac!)