# The curve $ay^{2}=x^{2}(3a-x)$ cuts the $y$-axis at

$\begin{array}{1 1}(1)x=-3a,x=0&(2)x=0,x=3a\\(3)x=0,x=a&(4)x=0\end{array}$

$ay^2 =x^2 (3a-x)$
The point of intersection with y-axis is obtained by putting $x=0$ in the equation of the curve.
$ay^2 =0 (3a -0)$
$y^2 =0 => y=0$
The curve intersect the y-axis at the origin ie $x=0$
Hence 4 is the correct answer.