Determine the minimum area of a parallelogram each of whose four sides are tangent to the ellipse x^{2}/49 + y^{2}/36 = 1.

(In reply to

Solution by Brian Smith)

The set of parallelograms which are tangent to the given ellipse all have something in common. That is the vertecies all lie on the same ellipse x^2/98 + y^2/72 = 1.

To show this, use the transform in my previous comment, but also add a circle which circumscribes the square. When inverting the transform, the circle becomes an ellipse.