Answer
If $c \gt \frac{1}{2},~~$ there are a total of three lines through the point $(0,c)$ that are normal to the parabola $y = x^2$
If $c \leq \frac{1}{2},~~$ there is one line through the point $(0,c)$ that is normal to the parabola $y = x^2$.
Work Step by Step
$y = x^2$
$\frac{dy}{dx} = 2x$
Note that the value of $\frac{dy}{dx}$ is the slope of the graph at each point $x$
At $x = 0$, the slope of the graph is $\frac{dy}{dx} = 2(0) = 0$
Therefore, the y-axis is normal to the parabola $y = x^2$ and the y-axis passes through every point which has the form $(0,c)$
Thus, for all points $(0,c)$, at least one line through the point $(0,c)$ is normal to the parabola.
We can find additional lines through the point $(0,c)$ that are normal to the parabola $y = x^2$
Suppose a line through the point $(0,c)$ is normal to the parabola. Then the slope of the line is $~~m = -\frac{1}{2x}$
The equation of the line is: $~~y-c = -\frac{1}{2x}(x-0)$
Then:
$y-c = -\frac{1}{2x}(x-0)$
$x^2-c = -\frac{1}{2x}(x)$
$x^2 = c-\frac{1}{2}$
$x = \pm\sqrt{c-\frac{1}{2}}$
If $c \gt \frac{1}{2}$, there are two additional solutions. If $c \leq \frac{1}{2}$, there are no additional solutions.
Therefore, if $c \gt \frac{1}{2},~~$ there are a total of three lines through the point $(0,c)$ that are normal to the parabola $y = x^2$
If $c \leq \frac{1}{2},~~$ there is one line through the point $(0,c)$ that is normal to the parabola $y = x^2$. Note that this line is the y-axis.
