Find the points on the curve `y=x^3-3x` , where the tangent to the curve is parallel to the chord joining `(1,\ -2)` and `(2,\ 2)`

Answer

`(x_{1}, y_{1})=(-sqrt{frac{7}{3}}, frac{2}{3} sqrt{frac{7}{3}})`

Explanation

Let `(x_{1}, y_{1})` 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 `= (\frac{d y}{d x})_{(x_{1}, y_{1})}=3 x_{1}^{2}-3`


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

`\therefore` Slope of the tangent `=` Slope of the chord
$$
\begin{aligned}
...