Question

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

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

Answer 1

Answer: (C)

$\frac{4a^2}{3}(4\pi +\sqrt{3})$

Solving the two, we get $x^2+6a-16a^2=0$

or $(x+8a)(x-2a)=0$

$\therefore x=2a$

$\therefore y=\pm 2\sqrt{3a}$

The required common area $=2\left[APOA\right]$

$=2\int ydx+2\int ydx$

$=2{\int }_0^{2a}\sqrt{6a}\sqrt{x}dx+2{\int }_{2a}^{4a}\sqrt{{\left(4a\right)}^2-x^2}dx$

$=2.\sqrt{6a}\frac{2}{3}{\left[x^{3/2}\right]}_0^{2a}+2.{[\frac{x}{2}\sqrt{{\left(4a\right)}^2-x^2}+\frac{1}{2}{\left(4a\right)}^2{\sin }^{-1}\frac{x}{4a}]}_{2a}^{4a}$

$=2.\sqrt{6a}\frac{2}{3}\left(2a\right)\sqrt{2a}+2[(0-a2\sqrt{3a})+8a^2({\sin }^{-1}1-{\sin }^{-1}\frac{1}{2})]$

$=2.2\sqrt{3}.\frac{4}{3}{a}^2+2[-2\sqrt{3}{a}^2(\frac{\pi }{2}-\frac{\pi }{6})]$

$=\frac{16}{3}\sqrt{3}{a}^2-4\sqrt{3}{a}^2+16a^2\frac{\pi }{3}$

$=\frac{4\sqrt{3}{a}^2}{3}+\frac{16\pi a^2}{3}=\frac{4a^2}{3}(4\pi +\sqrt{3})$