JEE Mains · Maths · STD 12 - 8. Application and integration
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.
- A \(\frac{1}{3}(2-12 \sqrt{3}+8 \pi)\)
- B \(\frac{1}{3}(2-12 \sqrt{3}+6 \pi)\)
- C \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
- D \(\frac{1}{3}(4-12 \sqrt{3}+6 \pi)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{3}(4-12 \sqrt{3}+8 \pi)\)
Step-by-step Solution
Detailed explanation
\(x^{2}+y^{2}+4 \sqrt{3} x-4=0\) \(y^{2}=8 x+4\) Point of intersections are \((0,2)\,and\, (0,-2)\) Both are symmetric about \(x\)-axis Area \(=2 \int_{0}^{2}\left(\sqrt{16-y^{2}}-2 \sqrt{3}\right)-\left(\frac{y^{2}-4}{8}\right)\) dy On solving Area…
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