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

Let a point on the ellipse be $P(a\cos \theta ,b\sin \theta )$.The rectangle inscribed in the ellipse will have vertices $(a\cos \theta ,b\sin \theta ),(-a\cos \theta ,b\sin \theta ),(-a\cos \theta ,-b\sin \theta )$, and $(a\cos \theta ,-b\sin \theta )$.The dimensions of the rectangle are:Length $=2a\cos \theta$ Breadth $=2b\sin \theta$ Area $A=(2a\cos \theta )(2b\sin \theta )=2ab(2\sin \theta \cos \theta )=2ab\sin 2\theta$ For the area to be maximum, $\sin 2\theta =1$.Thus, Maximum Area $=2ab(1)=2ab$.