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.

 

 

 

 
 
 
 
 

Answers (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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