# 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:

Let (x1y1) be the required point.

Slope of the chord $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2+2}{2-1}=4$

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

$\Rightarrow \frac{d y}{d x}=3 x^{2}-3 \ldots(1)$

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

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

$\therefore$ Slope of the tangent $=$ Slope of the chord

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

$\Rightarrow 3 x_{1}^{2}=7$

$\Rightarrow x_{1}^{2}=\frac{7}{3}$

$\Rightarrow x_{1}=\pm \sqrt{\frac{7}{3}}=\sqrt{\frac{7}{3}}$ or $-\sqrt{\frac{7}{3}}$

Case 1

When $x_{1}=\sqrt{\frac{7}{3}}$

On substituting this in eq. (1), we get

$y_{1}=\left(\sqrt{\frac{7}{3}}\right)^{3}-3\left(\sqrt{\frac{7}{3}}\right)=\frac{7}{3} \sqrt{\frac{7}{3}}-3 \sqrt{\frac{7}{3}}=\frac{-2}{3} \sqrt{\frac{7}{3}}$

$\therefore\left(x_{1}, y_{1}\right)=\left(\sqrt{\frac{7}{3}}, \frac{-2}{3} \sqrt{\frac{7}{3}}\right)$

Case 2

When $x_{1}=-\sqrt{\frac{7}{3}}$

On substituting this in eq. (1), we get

$y_{1}=\left(-\sqrt{\frac{7}{3}}\right)^{3}-3\left(-\sqrt{\frac{7}{3}}\right)=\frac{-7}{3} \sqrt{\frac{7}{3}}+3 \sqrt{\frac{7}{3}}=\frac{2}{3} \sqrt{\frac{7}{3}}$

$\therefore\left(x_{1}, y_{1}\right)=\left(-\sqrt{\frac{7}{3}}, \frac{2}{3} \sqrt{\frac{7}{3}}\right)$