The ratio of the area of the region bounded by the curve $a^4{y}^2=(2a-x)x^5$  to the area of the circle whose radius is $a$ is given by

$A=2/a^2{\int }_0^{2a}{x}^{5/2}\sqrt{(2a-x)}dx$

Put  $x=2a{\sin }^2\theta \therefore dx=4a\sin \theta \cos \theta d\theta$
$\therefore A=\frac{2}{a^2}{\int }_0^{\pi /2}{\left(2a\right)}^{5/2}{\sin }^5\theta {\left(2a\right)}^{\frac{1}{2}}\cos \theta \left(4a\sin \theta \cos \theta \right)d\theta$
$$\begin{gathered} =\frac{8a.\left(8a^3\right)}{a^2}{\int }_0^{\pi /2}{\sin }^6\theta {\cos }^2\theta d\theta \\ =64a^2\frac{\left(5.3.1\right)1}{8.6.4.2}.\frac{\mathrm{\pi }}{2} \\ =\frac{5}{4}{\mathrm{\pi a}}^2 \end{gathered}$$      
      
Also B = area of circle of radius a   $=\pi a^2$    $\therefore \boxed{\frac{A}{B}=\frac{5}{4}}$