Question

The area common to the circle $x^2+y^2=16a^2$ and the parabola $y^2=6ax$ is

• A. $4a^2(8\pi -\sqrt{3})$
• B. $\frac{4a^2(4\pi +\sqrt{3})}{3}$
• C. $\frac{4a^2(4\pi +\sqrt{3})}{5}$
• D. None

Answer 1

The correct option is B $\frac{4a^2(4\pi +\sqrt{3})}{3}$



Solving $x^2+y^2=16a^2,y^2=6ax$
$x^2+6ax-16a^2=0\Rightarrow (x+8a)(x-2a)=0\Rightarrow x=-8a,2a$
If $x=2a$ then, $y^2=12a^2\Rightarrow y=2\sqrt{3}a$
Area = $2{\int }_0^{2\sqrt{3}a}(\sqrt{16a^2-y^2}-\frac{y^2}{6a})dy=2{[\frac{y}{2}\sqrt{16a^2-y^2}+8a^{2}si{n}^{-1}\frac{y}{4a}-\frac{y^3}{18a}]}_0^{2\sqrt{3}a}$
$=2[\sqrt{3}a.2a+8a^2.\frac{\pi }{3}-\frac{4}{\sqrt{3}{a}^2}]=2[8a^2\frac{\pi }{3}+\frac{2}{\sqrt{3}}{a}^2]=\frac{4a^2(4\pi +\sqrt{3})}{3}$