## Maximal Rectangle in Ellipse

Construct the maximal rectangle in an ellipse with axes-ratio r = a/b (|EF|/|GH|). The solution is the rectangle inscribed in the ellipse with the same (r) side-ratio r = |AB|/|BC|.

A quick proof can be given using the affinity f, which maps the unit circle to the given ellipse. This can be realized with a diagonal matrix diag(a/2, b/2). f maps the square to the maximal inscribed rectangle. This is seen from the corresponding exercise for the circle (see MaximalRectangle.html ) and the fact that affinities preserve area-ratios. Avoiding the use of matrices, one can argue with the preservetion of parallelity by affinities to show that MN is parallel to FG etc..

A similar argument shows that the minimal rectangle circumscribing an ellipse is the one having sides parallel to the axes of the ellipse, hence again the ratio is r = a/b = |AB|/|CD|. The minimal circumscribed and the maximal inscribed are similar. Since the affinities preserve the ratios along a line, the lengths of parallel sides of the circumscribed to the inscribed is the same as the one of the squares circumscribed/inscribed respectively in a circle. This being sqrt(2) we have that the sides of the inscribed maximal rectangle are sqrt(2)*a, sqrt(2)*b and the area is 2*a*b, a, b being respectively the axes of the ellipse: 2*a = |EF| and 2*b = |GH|. The file ParaCircumscribed.html handles a problem, inverse, in some sense, to the present one. This is the problem of inscribing/circumscribing ellipses that are mimimal/maximal in a given parallelogram.
More general one can ask which are the parallelograms inscribed in an ellipse and which of them is the maximal one. This leads to the study of the parallelograms with conjugate diameters which have all the same area E, equal to the area of the maximal rectangle inscribed in an ellipse. The subject is studied in the file ParaInscribedEllipse.html .