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 

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

\[\text { Hence }, y_1 = {x_1}^2 - 4 x_1 + 5 . . . \left( 1 \right)\]

\[\text { Now }, y = x^2 - 4x + 5 \]

\[ \therefore \frac{dy}{dx} = 2x - 4\]

\[\text { Now }, \]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =2 x_1 -4\]

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

\[ \therefore 2 x_1 - 4 = 2\]

\[ \Rightarrow 2 x_1 = 6\]

\[ \Rightarrow x_1 = 3\]

\[\text {Also }, \]

\[ y_1 = 9 - 12 + 5 = 2 [\text { From eq. } (1)]\]

\[\text { Thus, the required point is }\left( 3, 2 \right).\]