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.

\[\text { Slope of the chord } = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 + 2}{2 - 1} = 4\]

\[y = x^3 - 3x\]

\[ \Rightarrow \frac{dy}{dx} = 3 x^2 - 3 . . . \left( 1 \right)\]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) {{=3x}_1}^2 -3\]

\[\text { It is given that the tangent and the chord are parallel } .\]

\[\therefore \text { Slope of the tangent } = \text { Slope of the chord }\]

\[ \Rightarrow 3 {x_1}^2 - 3 = 4\]

\[ \Rightarrow 3 {x_1}^2 = 7\]

\[ \Rightarrow {x_1}^2 = \frac{7}{3}\]

\[ \Rightarrow x_1 = \pm \sqrt{\frac{7}{3}} = \sqrt{\frac{7}{3}} or - \sqrt{\frac{7}{3}}\]

\[\text { Case }1\]

\[\text { When }x_1 = \sqrt{\frac{7}{3}}\]

\[\text { On substituting this in eq. (1), we get }\]

\[ y_1 = \left( \sqrt{\frac{7}{3}} \right)^3 - 3\left( \sqrt{\frac{7}{3}} \right) = \frac{7}{3}\sqrt{\frac{7}{3}} - 3\sqrt{\frac{7}{3}} = \frac{- 2}{3}\sqrt{\frac{7}{3}} \]

\[ \therefore \left( x_1 , y_1 \right) = \left( \sqrt{\frac{7}{3}}, \frac{- 2}{3}\sqrt{\frac{7}{3}} \right)\]

\[\text { Case }2\]

\[\text { When }x_1 = - \sqrt{\frac{7}{3}}\]

\[\text { On substituting this in eq. (1), we get }\]

\[ y_1 = \left( - \sqrt{\frac{7}{3}} \right)^3 - 3\left( - \sqrt{\frac{7}{3}} \right) = \frac{- 7}{3}\sqrt{\frac{7}{3}} + 3\sqrt{\frac{7}{3}} = \frac{2}{3}\sqrt{\frac{7}{3}} \]

\[ \therefore \left( x_1 , y_1 \right) = \left( - \sqrt{\frac{7}{3}}, \frac{2}{3}\sqrt{\frac{7}{3}} \right)\]