Question

Find a point on the curve y = x³ − 3x where the tangent is parallel to the chord joining (1, −2) and (2, 2).

Answer 1

Let (x₁, y₁) be the required point.

$$\begin{gathered} \text{Slope of the chord}=\frac{y_2-y_1}{x_2-x_1}=\frac{2+2}{2-1}=4 \\ y=x^3-3x \\ \Rightarrow \frac{dy}{dx}=3x^2-3...\left(1\right) \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}{{\text{=3x}}_1}^2\text{-3} \\ \text{It is given that the tangent and the chord are parallel.} \\ \text{∴ Slope of the tangent = Slope of the chord} \\ \Rightarrow 3{x_1}^2-3=4 \\ \Rightarrow 3{x_1}^2=7 \\ \Rightarrow {x_1}^2=\frac{7}{3} \\ \Rightarrow x_1=\pm \sqrt{\frac{7}{3}}=\sqrt{\frac{7}{3}}\text{or}-\sqrt{\frac{7}{3}} \\ \text{Case}1 \\ \mathrm{When}{x}_1=\sqrt{\frac{7}{3}} \\ \text{On substituting this in eq. (1), we get} \\ {y}_1={\left(\sqrt{\frac{7}{3}}\right)}^3-3\left(\sqrt{\frac{7}{3}}\right)=\frac{7}{3}\sqrt{\frac{7}{3}}-3\sqrt{\frac{7}{3}}=\frac{-2}{3}\sqrt{\frac{7}{3}} \\ \therefore \left(x_1,y_1\right)=\left(\sqrt{\frac{7}{3}},\frac{-2}{3}\sqrt{\frac{7}{3}}\right) \\ \text{Case}2 \\ \mathrm{When}{x}_1=-\sqrt{\frac{7}{3}} \\ \text{On substituting this in eq. (1), we get} \\ {y}_1={\left(-\sqrt{\frac{7}{3}}\right)}^3-3\left(-\sqrt{\frac{7}{3}}\right)=\frac{-7}{3}\sqrt{\frac{7}{3}}+3\sqrt{\frac{7}{3}}=\frac{2}{3}\sqrt{\frac{7}{3}} \\ \therefore \left(x_1,y_1\right)=\left(-\sqrt{\frac{7}{3}},\frac{2}{3}\sqrt{\frac{7}{3}}\right) \end{gathered}$$