Question

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

Answer 1

Let $(x_1,y_1)$ be required points

slope = $-1/2$

slope of line perpendicular = 2

as $(x_1,y_1)$ lies on curve

$y_1=x_1^2-4x_1+5$---(1)

$\frac{dy}{dx}=2x_1-4=2$

$2x_1=6$

$x_1=3$

$y_1=3^2-4\times 3+5=9-12+5=2$

$\therefore$ point is (3,2)