# Find a point on the curve

Question:

Find a point on the curve $y=x^{3}-3 x$ where the tangent is parallel to the chord joining $(1,-2)$ and $(2,2)$.

Solution:

Given:

The curve $y=x^{3}-3 x$

First, we will find the Slope of the tangent

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}\right)-\frac{\mathrm{d}}{\mathrm{dx}}(3 \mathrm{x})$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{3}-1-3\left(\frac{\mathrm{dx}}{\mathrm{dx}}\right)$

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

The equation of line passing through $\left(x_{0}, y_{0}\right)$ and The Slope $m$ is $y-y_{0}=m\left(x-x_{0}\right)$.

so The Slope, $m=\frac{y-y_{0}}{x-x_{0}}$

The Slope of the chord joining $(1,-2) \&(2,2)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2-(-2)}{2-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4 \ldots(2)$

From (1) & (2)

$3 x^{2}-3=4$

$\Rightarrow 3 x^{2}=7$

$\Rightarrow x^{2}=\frac{7}{3}$

$\Rightarrow x=\pm \sqrt{\frac{7}{3}}$

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

$\Rightarrow y=x\left(x^{2}-3\right)$

$\Rightarrow y=\pm \sqrt{\frac{7}{3}}\left(\left(\pm \sqrt{\frac{7}{3}}\right)^{2}-3\right)$

$\Rightarrow y=\pm \sqrt{\frac{7}{3}}\left(\left(\frac{7}{3}-3\right)\right.$

$\Rightarrow y=\pm \sqrt{\frac{7}{3}}\left(\frac{-2}{3}\right)$

we know that, $(\pm x-)=\mp$

$\Rightarrow y=\mp\left(\frac{-2}{3}\right) \sqrt{\frac{7}{3}}$

Thus, the required point is $x=\pm \sqrt{\frac{7}{3}} \& y=\mp\left(\frac{-2}{3}\right) \sqrt{\frac{7}{3}}$