On the curve $xy^2 = 2a^3 , a > 0$ find all the points where the normal to the curve pass the origin.

Question

On the curve $xy^2 = 2a^3 , a > 0$ find all the points where the normal to the curve pass the origin.
My idea.
Use formula for normal. $$(y-y_0)=-\frac{1}{y'(x_0)}(x-x_0)$$ So we need $y'$. We will find it by using implicit differentiation. $$y^2+2xyy'=0$$ $$y'=\frac{-y}{2x}$$ We know that $x=0=y$. $$-y_0=\frac{-2x_0^2}{y_0}$$ $$y_0^2=2x_0^2$$ But we know that $y_0^2=\frac{2a^3}{x_0}$. $$\frac{2a^3}{x_0}=2x_0^2$$ $$x_0^3=a^3$$ This is solution for $x_0$, and by supstitution we know what is $y_0$ Is my work ok?