# Find the points on the curve

Question:

Find the points on the curve $y=x^{3}-2 x^{2}-2 x$ at which the tangent lines are parallel to the line $y=2 x-3$.

Solution:

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

Given:

$y=2 x-3$

$\therefore$ Slope of the line $=\frac{d y}{d x}=2$'

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

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

$\Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 x_{1}^{2}-4 x_{1}-2$

It is given that the tangent and the given line are parallel.

$\therefore$ Slope of the tangent $=$ Slope of the given line

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

$\Rightarrow 3 x_{1}^{2}-4 x_{1}-4=0$

$\Rightarrow 3 x_{1}^{2}-6 x_{1}+2 x_{1}-4=0$

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

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

$\Rightarrow x_{1}=2$ or $x_{1}=\frac{-2}{3}$

Case 1

When $x_{1}=2$

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

$y_{1}=8-8-4=-4$

$\therefore\left(x_{1}, y_{1}\right)=(2,-4)$

Case 2

When $x_{1}=\frac{-2}{3}$

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

$y_{1}=\frac{-8}{27}-\frac{8}{9}+\frac{4}{3}=\frac{-8-24+36}{27}=\frac{4}{27}$

$\therefore\left(x_{1}, y_{1}\right)=\left(\frac{-2}{3}, \frac{4}{27}\right)$