Q. Area of the greatest rectangle that can be inscribed in the ellipse $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $ is
Rajasthan PETRajasthan PET 2007
Solution:
Let the coordinates of the vertices of the rectangle are
$ A(a\cos \theta ,b\sin \theta ), $
$ B(-a\cos \theta ,b\sin \theta ), $
$ C(-a\cos \theta ,-b\sin \theta ) $ and $ D(a\cos \theta ,-b\sin \theta ) $ .
Then, the length of the rectangle
$ AB=2a\text{ }cos\theta $ and breadth $ AQ=2b\text{ }\sin \theta $
$ \therefore $ Area of rectangle, $ A=AB\times AD $
$ =2a\cos \theta .2b\sin \theta $
$ \Rightarrow $ $ A=2ab\sin 2\theta $ ...(i)
$ \therefore $ $ \frac{dA}{d\theta }=2\times 2ab\cos 2\theta $
Put for maxima or minima $ \frac{dA}{d\theta }=0 $
$ \Rightarrow $ $ \cos 2\theta =0 $
$ \Rightarrow $ $ 2\theta =\frac{\pi }{2} $
$ \Rightarrow $ $ \theta =\frac{\pi }{4} $
Now, $ \frac{{{d}^{2}}A}{d{{\theta }^{2}}}=-8ab\sin 2\theta $
At $ \theta =\frac{\pi }{4},\left( \frac{{{d}^{2}}A}{d{{\theta }^{2}}} \right)<0 $
$ \therefore $ At $ \theta =\frac{\pi }{4}, $ area will be maximum.
$ \therefore $ Maximum area of the rectangle $ =2ab\sin 2\left( \frac{\pi }{4} \right)=2ab $
Alternate Method From Eq. (i)
$ A=2ab\sin 2\theta $ Since max $ (\sin 2\theta )=1 $
$ \therefore $ $ {{A}_{\max }}=2ab $
