Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 78

For what point does the normal line to the parabola $y = x-x^2$ at the
point $(1,0)$ intersect the parabola a second time. Illustrate with a sketch.
Given: $y = x -x^2 \quad$ $P(1,0)$

Solving for the slope of the tangent line

$
\begin{equation}
\begin{aligned}
y & = x - x^2\\
\\
y' & = \frac{d}{dx}(x) - \frac{d}{dx}(x^2)\\
\\
y' & = 1 - 2x
\end{aligned}
\end{equation}
$


Let $y' = $ slope$(m_T)$ of the tangent line

$
\begin{equation}
\begin{aligned}
y' = m_T &= 1 - 2x
&& \text{Substitute value of } x\\
\\
m_T &= 1-2(1)
&& \text{Simplify the equation}\\
\\
m_T &= -1
\end{aligned}
\end{equation}
$


Solving for the slope of the normal line

$
\begin{equation}
\begin{aligned}
m_N &= \frac{-1}{m_T} && \text{Substitute value of the slope of the tangent line}\\
\\
& = \frac{-1}{-1}\\
\\
m_N &= 1
\end{aligned}
\end{equation}
$

Solving for the equation of the normal line

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

Equating the normal line and the parabola to find the point of intersection

$
\begin{equation}
\begin{aligned}
\text{Normal line } \qquad y &= x - 1\\
\\
\text{Parabola} \qquad y& = x-x^2
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
x - x^2 &= x -1
&& \text{Add } -x \text{ to each sides}\\
\\
x - x - x^2 &= x - x - 1
&& \text{Comine like terms}\\
\\
-x^2 &= -1
&& \text{Multiply -1 to each sides}\\
\\
x^2 &= 1
&& \text{Take the square root of each sides}\\
\\
x =1, \quad x = -1
\end{aligned}
\end{equation}
$

Finding for the second point of intersection

$
\begin{equation}
\begin{aligned}
y &= x - x^2\\
y &= -1 - (-1)^2\\
y &= -2
\end{aligned}
\end{equation}
$


Thus, the point where the normal line intersects the parabola for the second time is at the point $(-1,-2,)$