Tangent and normal line to a parabola

Text

Given the parabola $y=x^{2}-4x+5$, find the equation of the tangent line and of the normal line at the point with abscissa $x_0=3$.

Solution

Ordinate: $y(3)=9-12+5=2$, so the point is $P(3,2)$. Derivative: $y'=2x-4$, hence the tangent slope is $m=y'(3)=2$. Tangent: $y-2=2(x-3)\Rightarrow y=2x-4$. Normal (slope $-\tfrac12$): $y-2=-\tfrac12(x-3)\Rightarrow y=-\tfrac12 x+\tfrac72$. $\boxed{\text{tangent } y=2x-4;\quad \text{normal } y=-\tfrac12 x+\tfrac72}$