The max area of a tangential parallelogram of an ellipse

Question

Given an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $a > b>0$ find out the minimum area of its tangential parallelogram, and specify when this max area is achieved.

Answer 1

Thanks to @Muralidharan, the map $(x,y)\rightarrow (x, \frac{by}{a})$ would solve it. The map maps an ellipse to a circle, and the tangential parallelograms will remain as the tangential parallelograms under this map. Thus the minimum is achieved when it is a rec-tangle tangent to a circle gets linearly transformed. Thus the area is $4ab$.

Plus I found a perfect way of visualizing the above. Let's have a two plane intersected with certain dihedral angle, and then orthogonal projection of a circle with tangential parallelograms onto another plane is just a ellipse with a tangential parallelogram. And we could calculate the area by the theorem of orthogonal projections of plane area regions