# Using integration, find the area of the greatest rectangle that can be inscribed in an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1.$

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$
$\Rightarrow y= \frac{b}{a}\sqrt{a^{2}-x^{2}}---(i)$
Let ABCD be the rectangle
Also $AD= 2x,AB= 2y$
Now area, $A= 4xy$
$\Rightarrow A^{2}= 16x^{2}y^{2}$
$\Rightarrow A^{2}= 16x^{2}\, \frac{b^{2}}{a^{2}}\left ( a^{2}-x^{2} \right )$
$\therefore f\left ( x \right )= \frac{16b^{2}}{a^{2}}\left ( a^{2}x^{2} -x^{4}\right )$  where $A^{2}= f\left ( x \right )$
$\Rightarrow {f}'\left ( x \right )= \frac{16b^{2}}{a^{2}}\left ( 2a^{2}x-4x^{3} \right )$
and  $\Rightarrow {f}'\left ( x \right )= \frac{32b^{2}}{a^{2}}\left ( a^{2}-6x^{2} \right )$
For $\Rightarrow {f}'\left ( x \right )= 0,\frac{16b^{2}}{a^{2}}\left ( 2a^{2}x-4x^{3} \right )= 0$
$\Rightarrow x= \frac{a}{\sqrt{2}}\; \; so\: y= \frac{b}{\sqrt{2}}\left [ from(i) \right ]$
$\because {f}''\left ( \frac{a}{\sqrt{2}} \right )= \frac{32b^{2}}{a^{2}}\left [a ^{2}-6\times \frac{a^{2}}{2} \right ]= -64b^{2}< 0$
$\therefore f\left ( x \right )$  is maximum and thus, it is also maximum at $x= \frac{a}{\sqrt{2}}$
Now required area   $=$ or $\left ( OEAFO \right )= 4\int_{0}^{\frac{a}{\sqrt{2}}}y \, dx$
$\Rightarrow 4\int_{0}^{\frac{a}{\sqrt{2}}}\frac{b}{\sqrt{2}}\, dx$
$\Rightarrow 2\sqrt{2}b\left [ x \right ]^{\frac{a}{\sqrt{2}}}_{0}= 2\sqrt{2}b\left [ \frac{a}{\sqrt{2}}-0 \right ]$
$=$ 2ab sq.units

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