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 $=4\underset{0}{\overset{9/2}{\int }}ydx$
$=4\underset{0}{\overset{9/2}{\int }}\frac{7}{3}\sqrt{1-\frac{x^2}{{\left(9/2\right)}^2}}dx$
$=\frac{28}{3}\underset{0}{\overset{9/2}{\int }}\frac{2}{9}\sqrt{{\left(\frac{9}{2}\right)}^2-x^2}dx$
$=\frac{56}{27}{\left[\frac{x}{2}\sqrt{\frac{81}{4}-x^2}+\frac{81}{4\times 2}si{n}^{-1}\left(\frac{2x}{9}\right)\right]}_0^{9/2}$
$=\frac{56}{27}\times \frac{81}{8}\times \frac{\pi }{2}=32.97$ sq. units