Question

Write the coordinates of the point at which the tangent to the curve y = 2x² − x + 1 is parallel to the line y = 3x + 9.

Answer 1

Let (x₁, y₁) be the required point.
Slope of the given line = 3
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=2{x_1}^2-x_1+1...\left(1\right) \\ \mathrm{Now},y=2x^2-x+1 \\ \therefore \frac{dy}{dx}=4x-1 \\ \mathrm{Now}, \\ \mathrm{S}\text{lope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}4x_1-1 \\ \text{Given:} \\ \text{Slope of the tangent = Slope of line} \\ \therefore 4x_1-1=3 \\ \Rightarrow x_1=1 \\ \text{From (1), we get} \\ {y}_1=2-1+1=2 \\ \therefore \left(x_1,y_1\right)=\left(1,2\right) \end{gathered}$$