Question

What will be the area enclosed by the ellipse given by $ \frac{4x^{2}}{81}+\frac{9y^{2}}{49}=1 $ ?

• A. $ 16.49 $ sq. units
• B. $ 24.72 $ sq. units
• C. $ 32.97 $ sq.units
• D. $ 48 $ sq. units

Answer 1

Answer: (C)

The given equation of ellipse is.

$\frac{x^{2}}{\left(\frac{9}{2}\right)^{2}} + \frac{y^{2}}{\left(\frac{7}{3}\right)^{2}} = 1 $
$ \therefore a = \frac{9}{2}$ and $ b = \frac{7}{3}$
$ \therefore $ Required area $= 4 \int\limits_{0}^{9/2} y dx $
$ = 4 \int\limits_{0}^{9/2} \frac{7}{3} \sqrt{1-\frac{x^{2}}{\left(9/2\right)^{2}}} dx $
$= \frac{28}{3} \int\limits_{0}^{9/2} \frac{2}{9} \sqrt{\left(\frac{9}{2}\right)^{2}-x^{2}}dx$
$ = \frac{56}{27} \left[\frac{x}{2}\sqrt{\frac{81}{4}-x^{2}} + \frac{81}{4\times2} sin^{-1} \left(\frac{2x}{9}\right)\right]_{0}^{9/2}$
$ = \frac{56}{27} \times\frac{81}{8}\times\frac{\pi}{2} = 32.97$ sq. units