Question

The area of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is :

• A. 6
• B. 36 $\pi$
• C. 36
• D. 6 $\pi$

Answer 1

Answer: (D)

Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$
$\Rightarrow 4x^2+9y^2=36$
$\Rightarrow 9y^2=36-4x^2$
$\Rightarrow y^2=4-\frac{4}{9}{x}^2$
$\Rightarrow y=\sqrt{4-\frac{4}{9}{x}^2}$
$\Rightarrow y=\frac{2}{3}\sqrt{9-x^2}$
Here, $a=3$, $b=4$
So, the required area (along x-axis)
$=4\underset{0}{\overset{3}{\int }}\frac{2}{3}\sqrt{9-x^2}dx=\frac{8}{3}\underset{0}{\overset{3}{\int }}\sqrt{{\left(3\right)}^2-x^2}dx$
$=\frac{8}{3}{\left[\frac{1}{2}x\sqrt{9-x^2}+\frac{9}{2}si{n}^{-1}\left(\frac{x}{3}\right)\right]}_0^3$
$\left(\begin{array}{c}\text{Using Formula}\int \sqrt{a^2-x^2}dx\\ =\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^{2}si{n}^{-1}\frac{x}{a}\end{array}\right)$
$=\frac{8}{3}\left[\frac{9}{2}si{n}^{-1}\left(\frac{3}{5}\right)-\frac{9}{2}si{n}^{-1}\left(0\right)\right]$
$=\frac{8}{3}\left[\frac{9}{2}.\frac{\pi }{2}\right]=6\pi sq$. units