Question

At what points on the curve y = 2x² − x + 1 is the tangent parallel to the line y = 3x + 4?

Answer 1

Let (x₁, y₁) be the required point.
The slope of line y = 3x + 4 is 3.

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence, y }_1 = 2 {x_1}^2 - x_1 + 1\]

\[\text { Now, y } = 2 x^2 - x + 1\]

\[\frac{dy}{dx} = 4x - 1\]

\[\text { Now,}  \]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =4 x_1 -1\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \text { Slope of the given line [Given] }\]

\[ \therefore 4 x_1 - 1 = 3\]

\[ \Rightarrow 4 x_1 = 4\]

\[ \Rightarrow x_1 = 1\]

\[\text { and }\]

\[ y_1 = 2 {x_1}^2 - x_1 + 1 = 2 - 1 + 1 = 2\]

\[\text { Thus, the required point is }\left( 1, 2 \right).\]