The area bounded by the curve $x{y}^2=a^2(a-x),a>0$ and $y-$axis is

The correct option is B $\pi a^2$ sq. units

$x{y}^2=a^2(a-x)$
$\Rightarrow x=\frac{a^3}{y^2+a^2}$
The given curve is symmetrical about $x-$axis and meets it at $(a,0).$
The line $x=0$ i.e., $y-$axis is an asymptote
$\therefore$ Area $=2\underset{0}{\overset{\infty }{\int }}xdy=2\underset{0}{\overset{\infty }{\int }}\frac{a^3}{y^2+a^2}dy$
$=2a^3\frac{1}{a}{\left[{\tan }^{-1}\frac{y}{a}\right]}_0^{\infty }$
$=2a^2\frac{\pi }{2}=\pi a^2$ sq. units