# Find the points on the curve

Question:

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

Solution:

Let $\left(x_{1}, y_{1}\right)$ represent the required points.

The slope of the $x$-axis is 0 .

Here,

$2 a^{2} y=x^{3}-3 a x^{2}$

Since, the point lies on the curve.

Hence, $2 a^{2} y_{1}=x_{1}^{3}-3 a x_{1}^{2} \quad \ldots$ (1)

Now, $2 a^{2} y=x^{3}-3 a x^{2}$

On differentiating both sides w.r.t. $x$, we get

$2 a^{2} \frac{d y}{d x}=3 x^{2}-6 a x$

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

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{3 x_{1}^{2}-6 a x_{1}}{2 a^{2}}$

Given:

Slope of the tangent at $\left(x_{1}, y_{1}\right)=$ Slope of the $x$-axis

$\Rightarrow \frac{3 x_{1}^{2}-6 a x_{1}}{2 a^{2}}=0$

$\Rightarrow 3 x_{1}^{2}-6 a x_{1}=0$

$\Rightarrow x_{1}\left(3 x_{1}-6 a\right)=0$

$\Rightarrow x_{1}=0$ or $x_{1}=2 a$

Also,

$2 a^{2} y_{1}=0$ or $2 a^{2} y_{1}=8 a^{3}-12 a^{3} \quad$ [From eq. (1)]

$\Rightarrow y_{1}=0$ or $y_{1}=-2 a$

Thus, the required points are $(0,0)$ and $(2 a,-2 a)$.