Question

Find the points on the curve y = x³ − 2x² − 2x at which the tangent lines are parallel to the line y = 2x− 3 ?

Answer 1

Let (x₁, y₁) be the required point.
Given:

\[y = 2x - 3\]

\[ \therefore \text { Slope of the line }= \frac{dy}{dx} = 2\]

\[y = x^3 - 2 x^2 - 2x\]

\[\text { Since } \left( x_1 y_1 \right) \text { lies on curve }, y_1 = {x_1}^3 - 2 {x_1}^2 - 2 x_1 . . . \left( 1 \right)\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = 3 {x_1}^2 - 4 x_1 - 2\]

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

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

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

\[ \Rightarrow 3 {x_1}^2 - 4 x_1 - 4 = 0\]

\[ \Rightarrow 3 {x_1}^2 - 6 x_1 + 2 x_1 - 4 = 0\]

\[ \Rightarrow 3 x_1 \left( x_1 - 2 \right) + 2 \left( x_1 - 2 \right) = 0\]

\[ \Rightarrow \left( x_1 - 2 \right) \left( 3 x_1 + 2 \right) = 0\]

\[ \Rightarrow x_1 = 2 or x_1 = \frac{- 2}{3}\]

\[\text { Case1 }\]

\[\text { When }x_1 = 2\]

\[\text { On substituting the value of } x_1 \text { in eq. (1), we get}\]

\[ y_1 = 8 - 8 - 4 = - 4\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 2, - 4 \right)\]

\[\text { Case 2}\]

\[\text { When }x_1 = \frac{- 2}{3}\]

\[\text { On substituting the value of } x_1 \text { in eq. (1), we get }\]

\[ y_1 = \frac{- 8}{27} - \frac{8}{9} + \frac{4}{3} = \frac{- 8 - 24 + 36}{27} = \frac{4}{27}\]

\[ \therefore \left( x_1 , y_1 \right) = \left( \frac{- 2}{3}, \frac{4}{27} \right)\]