Question

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

• A. $2ab$
• B. $ab$
• C. $\sqrt{ab}$
• D. $\frac{a}{b}$

Answer 1

Answer: (A)

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
$AB=2acos\theta$ and breadth $AQ=2b\sin \theta$
$\therefore$ 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 $(\sin 2\theta )=1$
$\therefore$ $A_{\max }=2ab$