Explanation

(c)
Given equation of curves are
$x^{2} + y^{2} = 16a^{2}$ and $y^{2} = 6ax$
The point of intersection are $x = 2a,\ y = \pm 2\sqrt{3}a$
[image:8_image81.jpeg]
$\therefore$ Required area,
$A = 2$ area of curve $APOP$
$= 2$[area of curve $OMPO +$area of curve $MAPM$]
$= 2\left\lbrack \int_{0}^{2a}{\sqrt{6a}\sqrt{x}dx} \right\rbrack + 2\left\lbrack \int_{2a}^{4a}{\sqrt{(4a)^{2} - x^{2}}dx} \right\rbrack$
$= 2 \bullet \frac{2}{3}\sqrt{6a}\ \left\lbrack x^{3/2} \right\rbrack_{0}^{2a} + 2\left\lbrack \frac{x}{2}\sqrt{(4a)^{2} - x^{2}} + \frac{1}{2}(4a)^{2}\sin^{- 1}\frac{x}{4a} \right\rbrack_{2a}^{4a}$
$= \frac{4}{3}\sqrt{6a}(2a)^{3/2} + 2\left\lbrack \left( 0 - 2a\sqrt{3a} \right) + 8a^{2}\left( \sin^{- 1}{1 - \sin^{- 1}\frac{1}{2}} \right) \right\rbrack$
$= \frac{16}{3}\sqrt{3}a^{2} - 4\sqrt{3}a^{2} + 16a^{2}\frac{\pi}{3}$
$\Rightarrow \ A = \frac{4\sqrt{3}a^{2}}{3} + \frac{16\pi a^{2}}{3}$
$= \frac{4a^{2}}{3}(4\pi + \sqrt{3})\ sq\ unit$