Find the points on the curve

`
Question:

Find the points on the curve $y^{2}=2 x^{3}$ at which the slope of the tangent is 3 .

Solution:

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

Given:

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

Since $\left(x_{1} y_{1}\right)$ lies on a curve, $y_{1}^{2}=2 x_{1}{ }^{3}$      .....(i)

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

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

Slope of the tangent at $(x, y)=\frac{3 x_{1}{ }^{2}}{y_{1}}$

Slope of the tangent $=3$    [Given]

$\therefore \frac{3 x_{1}^{2}}{y_{1}}=3$             ....(2)

$\Rightarrow y_{1}=x_{1}^{2}$

On substituting the value of $y_{1}$ in eq. (1), we get

$x_{1}^{4}=2 x_{1}^{3}$

$\Rightarrow x_{1}^{3}\left(x_{1}-2\right)=0$

$\Rightarrow x_{1}=0,2$

Case 1

When $x_{1}=0, y_{1}=x^{2}=0$. Thus, we get the point $(0,0)$. But, it does not satisfy eq. (2).

So, we can ignore $(0,0)$.

Case 2

When $x_{1}=2, y_{1}=x_{1}^{2}=4$. Thus, we get the point $(2,4)$.

Leave a comment