Find the points on the curve `2a^2y=x^3-3a x^2` where the tangent is parallel to x-axis.

Answer

The required points are `(0,0)` and `(2 a,-2 a)`

Explanation

Let `(x_{1}, y_{1})` be the required points.

The slope of the `x`-axis is 0

Here `2 a^{2} y=x^{3}-3 a x^{2}`,

`2 a^{2} y_{1}=x_{1}^{3}-3 a x_{1}^{2} (1)`


Now `2 a^{2} y=x^{3}-3 a x^{2}`

On differentiating both sides wrt `x`,
$$
\begin{aligned}
...