Find the point on the curve

Given:

The curve is $2 a^{2} y=x^{3}-3 a x^{2}$

Differentiating the above w.r.t $\mathrm{x}$

$\Rightarrow 2 \mathrm{a}^{2} \times \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{3}-1-3 \times 2 \mathrm{ax}^{2-1}$

$\Rightarrow 2 \mathrm{a}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-6 \mathrm{ax}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 \mathrm{x}^{2}-6 \mathrm{ax}}{2 \mathrm{a}^{2}} \ldots(1)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=$ The Slope of the tangent $=\tan \theta$

Since, the tangent is parallel to $x$-axis

i.e,

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\tan (0)=0$      .....(2)

$\therefore \tan (0)=0$

$\therefore \frac{d y}{d x}=$ The Slope of the tangent $=\tan \theta$

From (1) & (2), we get,

$\Rightarrow \frac{3 x^{2}-6 a x}{2 a^{2}}=0$

$\Rightarrow 3 x^{2}-6 a x=0$

$\Rightarrow 3 x(x-2 a)=0$

$\Rightarrow 3 x=0$ or $(x-2 a)=0$

$\Rightarrow x=0$ or $x=2 a$

Substituting $x=0$ or $x=2 a$ in $2 a^{2} y=x^{3}-3 a x^{2}$,

when $x=0$

$\Rightarrow 2 a^{2} y=(0)^{3}-3 a(0)^{2}$

$\Rightarrow y=0$

when $x=2$

$\Rightarrow 2 a^{2} y=(2 a)^{3}-3 a(2 a)^{2}$

$\Rightarrow 2 a^{2} y=8 a^{3}-12 a^{3}$

$\Rightarrow 2 a^{2} y=-4 a^{3}$

$\Rightarrow y=-2 a$

Thus, the required point is $(0,0) \&(2 a,-2 a)$