Question

The area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

• A. $12\pi$
• B. $3\pi$
• C. $24\pi$
• D. $\pi$

Answer 1

Answer: (A)

The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$
The given ellipse is symmetrical about both axis as it contains only even powers of $y$ and $x$.

Now, $\frac{x^2}{16}+\frac{y^2}{9}=1\Rightarrow \frac{y^2}{9}=1-\frac{x^2}{16}\Rightarrow y^2=\frac{9}{16}\left(16-x^2\right)$
$\Rightarrow y=\pm \frac{3}{4}\left(\sqrt{16-x^2}\right)$
Now, area bounded by the ellipse $=4$ (area of ellipse in first quardant)
$=4($ area $OAC)$
$=4\underset{0}{\overset{4}{\int }}ydx=\underset{0}{\overset{4}{\int }}\frac{3}{4}\sqrt{16-x^2}dx[\because y\ge 0$ in first quadrant $]$
Put $x=4\sin \theta$ so that $dx=4\cos \theta d\theta$
Now when $x=0,\theta =0$ and when $x=4,\theta =\frac{\pi }{2}$
$\therefore$ Req. area $=\frac{4\times 3}{4}{\int }_0^{\frac{\pi }{2}}\sqrt{16-16{\sin }^2}\theta \cdot 4\cos \theta d\theta$
$=3\underset{0}{\overset{\frac{\pi }{2}}{\int }}4\sqrt{1-{\sin }^2\theta }\cdot 4\cos \theta d\theta$
$=48\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^2\theta d\theta =48{\int }_0^{\frac{\pi }{2}}\left(\frac{1+\cos 2\theta }{2}\right)d\theta$
$=24\underset{0}{\overset{\frac{\pi }{2}}{\int }}(1+\cos 2\theta )d\theta =24{\left[\theta +\frac{\sin 2\theta }{2}\right]}_0^{\frac{\pi }{2}}$
$=24\left[\left(\frac{\pi }{2}-0\right)+\frac{1}{2}(0-0)\right]=12\pi$ sq. units.