Mathematics · Area under the Curves

JEE Main 2024 — 9 April, Shift 2 — Question 5

The area (in square units) of the region enclosed by the ellipse $x^{2}+3 y^{2}=18$ in the first quadrant below the line $\mathrm{y}=\mathrm{x}$ is :

Answer: C

$\frac{x^{2}}{18}+\frac{y^{2}}{6}=1$

$\frac{x^{2}}{18}+\frac{3 x^{2}}{18}=1 \Rightarrow 4 x^{2}=18 \Rightarrow x^{2}=\frac{9}{2}$

$\int_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \frac{\sqrt{18-\mathrm{x}^{2}}}{\sqrt{3}} d x$

$=\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^{2}}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}}$

$=\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right)$

Required Area $=\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}}$ $=\sqrt{3} \pi$

Answer key and solution verified before publishing.

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