Question

Find the area of the greatest rectangle that can be inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Answer 1

Let $(a\cos \theta ,b\sin \theta )$ be one of the vertices of the rectangle.

Hence the length$=2a\cos \theta$ and the width$=2b\sin \theta$

$\Rightarrow$Area of the rectangle$A=2a\cos \theta \times 2b\sin \theta =4ab\sin \theta \cos \theta =2ab\sin 2\theta$

$A^{\prime }=2ab\times 2\cos 2\theta$

For maximum or minimum values of $A$

$\Rightarrow A^{\prime }=0$

$\Rightarrow 4ab\cos 2\theta =0$

$\Rightarrow \cos 2\theta =\frac{\pi }{2}$

$\Rightarrow 2\theta =\frac{\pi }{2}$

$\Rightarrow \theta =\frac{\pi }{4}$

$\Rightarrow A=2ab\sin \left(\frac{2\pi }{4}\right)$

$\Rightarrow A=2ab\sin \left(\frac{\pi }{2}\right)$

$\therefore A=2ab$

Now $A^{\prime \prime }=-8ab\sin 2\theta$

Area is maximum when $A=2ab$