Question

At what points on the curve y = x² − 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Answer 1

Let (x₁, y₁) be the required point.
Slope of the given line = $\frac{-1}{2}$
Slope of the line perpendicular to this line = 2

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1={x_1}^2-4x_1+5...\left(1\right) \\ \mathrm{Now},y=x^2-4x+5 \\ \therefore \frac{dy}{dx}=2x-4 \\ \mathrm{Now}, \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=2}{\mathit{\text{x}}}_1\text{-4} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=Slope of the given line [Given]} \\ \therefore 2x_1-4=2 \\ \Rightarrow 2x_1=6 \\ \Rightarrow x_1=3 \\ \mathrm{Also}, \\ {y}_1=9-12+5=2[\text{From eq. (1)]} \\ \text{Thus, the required point is}\left(3,2\right)\text{.} \end{gathered}$$