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Question

Mathematics Question on Area between Two Curves

The area of the smaller region enclosed by the curves y 2 = 8 x + 4 and
x2+y2+4√3x-4=0
is equal to

A

13(2−12√3+8π)\frac{1}{3}(2-12√3+8π)

B

13(2−12√3+6π)\frac{1}{3}(2-12√3+6π)

C

13(4−12√3+8π)\frac{1}{3}(4-12√3+8π)

D

13(4−12√3+6π)\frac{1}{3}(4-12√3+6π)

Answer

13(4−12√3+8π)\frac{1}{3}(4-12√3+8π)

Explanation

Solution

The correct answer is (C):

The area of the smaller region enclosed by the curves y2 = 8x + 4
cosθ=2√34cosθ =\frac{ 2√3}{4}
= √3/4
⇒ θ = 30°
Area of the required region
=23(4×12)+42×π6−12×4×2√3\frac{ 2}{3}(4×\frac{1}{2})+42×\frac{π}{6}-\frac{1}{2}×4×2√3
= 43+8π3−4√3\frac{4}{3}+\frac{8π}{3}-4√3
= 134−12√3+8π\frac{1}{3}{4-12√3+8π}