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 $(-1,-2,)$