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)

6 $\pi$

Answer 2

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) $= 4\int\limits^{3}_{0} \frac{2}{3}\sqrt{9-x^{2}} dx = \frac{8}{3} \int\limits^{3}_{0}\sqrt{\left(3\right)^{2}-x^{2}} dx$ $= \frac{8}{3}\left[\frac{1}{2}x\sqrt{9-x^{2}} + \frac{9}{2}sin^{-1}\left(\frac{x}{3}\right)\right]^{3}_{0}$ $\begin{pmatrix}\text{Using Formula}\,\int\sqrt{a^{2} - x^{2}}\,dx\\\ = \frac{1}{2}x\sqrt{a^{2} - x^{2}} + \frac{1}{2}a^{2} \,sin^{-1} \frac{x}{a}\end{pmatrix}$ $= \frac{8}{3}\left[\frac{9}{2}sin^{-1}\left(\frac{3}{5}\right) - \frac{9}{2}sin^{-1}\left(0\right)\right]$ $= \frac{8}{3}\left[\frac{9}{2} . \frac{\pi}{2}\right] = 6\pi\,sq$. units