Suppose there is a rectangle $ABCD$ with $A=(-2,1), B=(-2,-1), C=(2,-1), D=(2,1).$ Then we have an ellipse $\frac{x^2}{4} + y^2=1$ that is tangent to the rectangle. Then we build the rectangle $EFGH$ which has the corners as the intersection of the ellipse and the diagonals of $ABCD$ as can be seen in 
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The question is that how do we prove $\textbf{analytically}$ that $\frac{A(ABCD)}{A(EFGH)}=2$ for any ellipse?
I already know that it is really easy and cool to prove it by using the homeomorphism of the ellipse and the circle by deforming the question into the easier form.
Is this the analytical proof you are looking for?