Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 77

Determine the equation of the normal line to the parabola $y = x^2 - 5x + 4$ which is parallel to the line
$x - 3y = 5$


$\begin{equation} \begin{aligned} \text{Given:}&&& \text{Parabola}\quad & &y = x^2-5x+4\\ \phantom{x}&&& \text{Line} \quad & & x - 3y = 5 \end{aligned} \end{equation}$


Solving for slope$(m)$ using the equation of the line

$\begin{equation} \begin{aligned} x - 3y &= 5 && \text{Transpose } x \text{ to the right side}\\ \\ -3y &= 5-x && \text{Divide both sides by -3}\\ \\ \frac{-3y}{-3} &= \frac{5-x}{-3} && \text{Simplify the equation}\\ \\ y &= \frac{x-5}{3} \quad \text{ or } \quad y = \frac{1}{3}x - \frac{5}{3} && \text{By using the general formula of the equation of the line}\\ \\ y &= mx + b && \text{Slope}(m) \text{ is the numerical coefficient of }x\\ \\ m & = \frac{1}{3} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} y &= x^2 - 5x + 4\\ \\ y'&= \frac{d}{dx} (x^2) - 5 \frac{d}{dx} (x) = \frac{d}{dx}(4) && \text{Derive each term}\\ \\ y'&= 2x - (5) (1) + 0 && \text{Simplify the equation}\\ \\ y' &= 2x - 5 \end{aligned} \end{equation}$


Solving for the equation of the normal line,

$\begin{equation} \begin{aligned} m_N &= \frac{-1}{m} && \text{Slope of the normal line is equal to the negative reciprocal to the slope of the line which is } \frac{1}{3}\\ \\ m_N &= \frac{-1}{\frac{1}{3}}\\ \\ m_N &= -3 \end{aligned} \end{equation}$


Let $y' = m_N$

$\begin{equation} \begin{aligned} y' & = m_N = 2x - 5 && \text{Substitute the value of slope}(m_N)\\ \\ -3 & = 2x - 5 && \text{Add 5 to each sides}\\ \\ 2x &= 5 - 3 && \text{Combine like terms}\\ \\ 2x &= 2 && \text{Divide both sides by 2}\\ \\ \frac{2x}{2} &= \frac{2}{2} && \text{Simplify the equation} \end{aligned} \end{equation}$


Using equation of the parabola

$\begin{equation} \begin{aligned} y & = x^2 - 5x + 4 && \text{Substitute the value of } x\\ \\ y & = (1)^2 - 5(1)+4 && \text{Simplify the equation}\\ \\ y & = 0 \end{aligned} \end{equation}$


Using point slope form

$\begin{equation} \begin{aligned} y - y_1 &= m(x-x_1) && \text{Substitute value of } x,y \text{ and slope}(m)\text{ of the line}\\ \\ y - 0 &= \frac{1}{3} && \text{Simplify the equation} \end{aligned} \end{equation}$


The equation of the normal line is $\displaystyle y = \frac{x-1}{3}$