# The maximum area of a hexagon that can inscribed in an ellipse

I have to find the maximum area of a hexagon that can inscribed in an ellipse. The ellipse has been given as $\frac{x^2}{16}+\frac{y^2}{9}=1$.

I considered the circle with the major axis of the ellipse as the diameter of the circle with centre as the origin. By considering lines of $y=\pm\sqrt{3}x$, you can find the points on the circle which are vertices of a regular hexagon(including the end points of the diameter). I brought these points on the given ellipse such that the $x$ coordinate of the vertices remain the same and only the $y$ coordinate changes depending upon the equation of the ellipse. But that did not lead to a hexagon.

Since the question nowhere mentions of the hexagon being regular, I think it might be a irregular hexagon but any such hexagon can be drawn which encompasses most of the area of the ellipse.

• Possible hint. For the circle the answer is the regular hexagon. Then stretch by factors of $3$ and $4$ on the two axes. – Ethan Bolker May 10 '17 at 18:38
• Can you elaborate a little. The circle is already of radius 4 so stretching by a factor of 4 would exceed the ellipse – olympiad math May 10 '17 at 18:43
• Both answers show this approach, so there's no need to elaborate. – Ethan Bolker May 10 '17 at 18:55

We can use the affine transformation that $x=4u,y=3v$. Therefore, The ellipse will maps to the unit circle $u^2+v^2=1$. One of the most important property of affine thansformations is invariance of ratio of the areas. Ratio of areas of ellipse and unit circle is $4\cdot 3 = 12$.Now we have to find a cyclic hexagon that has maximum area. (By isoperimetric inequaliy, this hexagon must be regular).
Area of regular hexagon is $\dfrac{3\sqrt3}{2}$ and therefore, the hexagon has maximum area that in ellipse is $$12\cdot \dfrac{3\sqrt3}{2} =18\sqrt3$$