Explanation

(b)
Given curve $a^{4}y^{2} = (2a - x)x^{5}$
Cut off $x$-axis, when $y = 0$
$0 = (2a - x)x^{5}$
$\therefore x = 0,\ 2a$
Hence, the area bounded by the curve
$a^{4}y^{2} = (2a - x)x^{5}$ is
$A_{1} = \int_{0}^{2a}{\frac{\sqrt{(2a - x)}x^{5/2}}{a^{2}}dx}$
Put $x = 2a\sin^{2}\theta$
$\therefore dx = 4a\sin{\theta\cos{\theta d\theta}}$
$\therefore A_{1} = \int_{0}^{\pi/2}{\frac{\sqrt{2a}\cos{\theta(2a)^{5/2}\sin^{5}{\theta\ 4a\sin{\theta\cos\theta}}}}{a^{2}}d\theta}$
$= 32a^{2}\int_{0}^{\pi/2}{\sin^{6}{\theta\cos^{2}{\theta\ d\theta}}}$
$= 32a^{2} \bullet \frac{(5 \bullet 3 \bullet 1)(1)}{8 \bullet 6 \bullet 4 \bullet 2} \bullet \frac{\pi}{2}$ (by walli’s formula)
$= \frac{5\pi a^{2}}{8}$
Area of circle, $A_{2} = \pi a^{2}$
$\therefore\ \frac{A_{1}}{A_{2}} = \frac{5}{8}$
$\Rightarrow \ A_{1}:A_{2} = 5:8$