Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 101

How many lines through the point $(0,c)$ are normal lines to the parabola $y=x^2$ if $\displaystyle c > \frac{1}{2}$?
What if $\displaystyle \leq \frac{1}{2}$?

Recall that the slope of the normal line is equal to the negative reciprocal of the slope of the tangent line. So,


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


We can get the equation of the normal lines by using the slopes formula at the point of tangency
at $(x,x^2)$ and at $(0,c)$ and equate it with the slope of the normal line. So...


$\begin{equation} \begin{aligned} m_N &= \frac{y_2-y_1}{x_2-x_1}\\ \frac{-1}{2x} &= \frac{c-x^2}{0-x} && \text{(Applying cross multiplication)}\\ x &= 2xc - 2x^3\\ 2x^3 - 2xc + x &= 0\\ x(2x^2-2c+1) &= 0 \end{aligned} \end{equation}$


Its either $x=0$ and $2x^2-2c+1=0$


$\begin{equation} \begin{aligned} 2x^2 - 2c + 1 &= 0\\ \frac{\cancel{2}x^2}{\cancel{2}} &= \frac{2c-1}{2}\\ \sqrt{x^2} &= \sqrt{c - \frac{1}{2}}\\ x &= \pm \sqrt{c-\frac{1}{2}} \end{aligned} \end{equation}$


If $\displaystyle c > \frac{1}{2}$ let's say $c=2$, $\displaystyle x=\pm\sqrt{2 - \frac{1}{2}} \Longrightarrow x = + \frac{\sqrt{6}}{2}$ and $\displaystyle x = \frac{-\sqrt{6}}{2}$

You will have 2 normal lines. However, if $\displaystyle c \leq \frac{1}{2}$, let's say $\displaystyle c=\frac{1}{2}$
and $\displaystyle c = -\frac{1}{2}$, $\displaystyle x = \sqrt{\frac{1}{2}-\frac{1}{2}} = 0$ and
$\displaystyle x = \sqrt{-\frac{1}{2}- \frac{1}{2}} = \sqrt{-\frac{1}{4}}$, there will be only 1 normal line since
square root of a negative value is undefined.

Therefore,

$\begin{equation} \begin{aligned} & \text{if } c > \frac{1}{2}, \quad \text{3 normal lines} && (\text{including } x = 0 \text{ we've had here } x(2x^2-2c+1)=0 )\\ & \text{if } c \leq \frac{1}{2}, \quad \text{1 normal line} \end{aligned} \end{equation}$