The area of the smaller region enclosed by the curves y2 = 8x + 4 and
x2+y2+4√3x-4=0
is equal to
The area of the smaller region enclosed by the curves y2 = 8x + 4 and
x2+y2+4√3x-4=0
is equal to
- \(\frac{1}{3}(2-12√3+8π)\)
- \(\frac{1}{3}(2-12√3+6π)\)
- \(\frac{1}{3}(4-12√3+8π)\)
- \(\frac{1}{3}(4-12√3+6π)\)
The Correct Option is C
Solution and Explanation
The correct answer is (C):
\(cosθ =\frac{ 2√3}{4}\)
= √3/4
⇒ θ = 30°
Area of the required region
=\(\frac{ 2}{3}(4×\frac{1}{2})+42×\frac{π}{6}-\frac{1}{2}×4×2√3\)
= \(\frac{4}{3}+\frac{8π}{3}-4√3\)
= \(\frac{1}{3}{4-12√3+8π}\)
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Concepts Used:
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