Parallelogram inscribed in an ellipse

Start with a square with vertices $(0, 0), (1, 0), (1, 1), (0, 1)$ . We want to transform the square into a parallelogram with sides $a$ and $b$ , with $a \gt b$ and also the angle between the sides is $60^\circ$ .

For this we use the transformation:

$T =\begin{bmatrix} a && b \cos 60^{\circ} \\ 0 && b \sin 60^\circ \end{bmatrix}$

Now the minimum area ellipse passing through the vertices of the square we started with is a circle with center $C' = (\frac{1}{2}, \frac{1}{2})$ and radius $\dfrac{\sqrt{2}}{2}$ , so its equation is

$(r' - C')^T (r' - C') = \frac{1}{2}$

Now we apply the transformation to the circle, and for proper selection of $a$ and $b$ , this should gives us an ellipse with semi-axes $12$ and $6$ . Note that the ellipse generated has the minimum possible area for the resulting parallelogram, which is equivalent to saying the parallelogram has the the maximum possible area for the ellipse.

If $r$ is the transformed point, then $r = T r'$ and this implies that $r' = T^{-1} r$ . Substitute this into the equation of the circle, and this gives you the equation of the ellipse:

$(T^{-1} r - C')^T (T^{-1} r - C') = \frac{1}{2}$

which, after simple manipulation, becomes

$(r - T C')^t (2 T^{-t} T^{-1} ) (r - T C') = 1$

We want to find $a, b$ such the eigenvalues of $(2 T^{-t} T^{-1} )$ are the reciprocals of $12^2$ and $6^2$ ; or , such that the eigenvalues of $(2 T^{-t} T^{-1})^{-1}$ are $12^2$ and $6^2$

So,

$Q = (2 T^{-t} T^{-1})^{-1} = \dfrac{1}{2} T T^t$

hence,

$Q = \dfrac{1}{2} \begin{bmatrix} a && b \cos 60^\circ \\0 && b \sin 60^\circ \end{bmatrix} \begin{bmatrix} a && 0 \\ b \cos 60^\circ && b \sin 60^\circ \end{bmatrix}$

multiplying out and substituting the values of $\cos 60^\circ$ and $\sin 60^\circ$

$Q = \dfrac{1}{2} \begin{bmatrix} a^2 + \dfrac{b^2}{ 4} && b^2 \dfrac{\sqrt{3}}{4} \\ b^2 \dfrac{\sqrt{3}}{4} && b^2 \dfrac{3}{ 4} \end{bmatrix}$

The trace of this matrix is the sum of its eigenvalues , and its determinant is the product of its eigenvalues

hence, we want to have

$\frac{1}{2} ( a^2 + b^2 ) = 12^2 + 6^2$

and

$\frac{1}{4 } ( (a^2 + \dfrac{b^2}{4})(\dfrac{3 b^2}{4}) - b^4 (\dfrac{3}{16}) ) =( 12^2)( 6^2 )$

which simplifies nicely to

$\frac{3}{16} a^2 b^2 = (12^2 )( 6^2 )$

Let $u = a^2$ , $v = b^2$ , then

$u + v = 2( 144 + 36) = 2 (180) = 360$

and $u v = (144)(36)( 16) / 3 = 27648$

from the first of these two equations,

$v = 360 - u$

substitute this into the second equation

$u (360 - u) = 27648$

or $u^2 - 360 u + 27648 = 0$

hence

$u = \frac{1}{2} ( 360 + \sqrt{360^2 - 4(27648)} ) =248.934751758456343$

hence $a = \sqrt{u} = 15.777666232952715$

thus, the answer is $\boxed{15777666}$