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 $\displaystyle y = \frac{x-1}{3}$