Question

Area of the greatest rectangle that can be inscribed in an ellipse $9x^2+16y^2=144$ is equal to $($in sq. units$)$

Answer 1

Equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$
Let $P\equiv (4\cos \theta ,3\sin \theta )$
$\therefore PQ=6\sin \theta ,QR=8\cos \theta$
Area of rectangle, $f\left(\theta \right)=\left(8\cos \theta \right)\left(6\sin \theta \right)=24\sin 2\theta ,f^{\prime }\left(\theta \right)=48\cos 2\theta$
$f^{\prime \prime }\left(\theta \right)=-96\sin 2\theta$
Now, $f\left(\theta \right)$ is maximum when $\cos 2\theta =0$
$\Rightarrow \theta =\frac{\pi }{4}$
So, Maximum area of rectangle $=24$sq. units