QJEEM22Maa11e: JEE Main 2022 Mathematics - Class 12 JEEM22Maa11e Solution: The area of the smaller region enclosed by the curves and is…

Question

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The area of the smaller region enclosed by the curves $y^{2}=8 x+4$ and $x^{2}+y^{2}+4 \sqrt{3} x-4=0$ is equal to

Accepted Answer

JEE Main 2022 (Online) 27th July Morning Shift Mathematics - Area Under The Curves Question 63 English Explanation

$\cos \theta = {{2\sqrt 3 } \over 4} = {{\sqrt 3 } \over 2} \Rightarrow \theta = 30^\circ$

Area of the required region

$= {2 \over 3}\left( {4 \times {1 \over 2}} \right) + {4^2} \times {\pi \over 6} - {1 \over 2} \times 4 \times 2\sqrt 3$

$= {4 \over 3} + {{8\pi } \over 3} - 4\sqrt 3 = {1 \over 3}\left\{ {4 - 12\sqrt 3 + 8\pi } \right\}$

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