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  • The area of the smaller region enclosed by the curves  and <img alt="\mathrm{ x^{2}+y^{2

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

Option: 1

\mathrm{\frac{1}{3}(2-12 \sqrt{3}+8 \pi)}


Option: 2

\mathrm{\frac{1}{3}(2-12 \sqrt{3}+6 \pi)}


Option: 3

\mathrm{\frac{1}{3}(4-12 \sqrt{3}+8 \pi)}


Option: 4

\mathrm{\frac{1}{3}(4-12 \sqrt{3}+6 \pi)}


Answers (1)

best_answer

\mathrm{x^{2}+y^{2}+4 \sqrt{3} x-4=0} \\

\mathrm{y^{2}=8 x+4}

Point of intersections are (0,2) & (0,-2)both are symmetric about x-axis

\mathrm{\text { Area } =2 \int_{0}^{2}\left(\sqrt{16-y^{2}}-2 \sqrt{3}\right)-\left(\frac{y^{2}-4}{8}\right) d y} \\

            \mathrm={\frac{1}{3}[8 \pi+4-12 \sqrt{3}]}

Hence correct option is 3

Posted by

Sanket Gandhi

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