Mathematics · 2022
JEE Main · 27 July 2022, Shift 1 · Q9
The area of the smaller region enclosed by the curves y^2=8 x+4 and x^2+y^2+4 √ 3 x-4=0 is equal to
The area of the smaller region enclosed by the curves $\displaystyle y^2=8 x+4$ and $\displaystyle x^2+y^2+4 \sqrt{3} x-4=0$ is equal to
Official answer
(3)
$\displaystyle \frac{1}{3}(4-12 \sqrt{3}+8 \pi)$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.