How Many Lines Passing Through $(0, c>\frac{1}{2})$ Are Normal Lines To $y = x^2$ How Many Lines Passing Through $(0, c>\frac{1}{2})$ Are Normal Lines To $y = x^2$
What I've got so far:
Let $g$ be the line that intersects the parabola perpendicularly.
Let $P(p,p^2)$ be the points on the parabola that $g$ intersects perpendicularly.
Let $m_n$ be the slope of $g$.
Slope of the normal line: $(y'(p))(m_n) =(2p)(m_n) = -1\Leftrightarrow m_n = -\frac{1}{2p}$
So $g = -\frac{1}{2p}(x-p) + p^2$
It is given that the points $(x,g)$ can take on the values $(0,g)$ with $g>\frac{1}{2}$.
Substitute in the permissible values: $-\frac{1}{2p}(0-p) + p^2= \frac{1}{2}+p^2 > \frac{1}{2} \Leftrightarrow p^2> 0$.
And now I'm stuck.
Can anybody comment on what I've done so far and give me a hint regarding solving this problem?
 A: By the optical properties of the parabola, the tangent in $(x_0,x_0^2)$ cuts the $x=0$ line in the point $(0,-x_0^2)$, hence the normal cuts the $x=0$ line in the point $(0,x_0^2+1/2)$ by the Euclid's second theorem. So for any $c>1/2$ there are only two normals passing through $(0,c)$, namely the normal through $(\sqrt{c-1/2},c-1/2)$ and the normal through $(-\sqrt{c-1/2},c-1/2)$. Just the normal through the origin if $c=1/2$, no normals if $c<1/2$.
A: Given,
$$y=x^2$$
Let a point $(a, a^2)$ be there on the parabola,
Therefore,
$$f'(a) = 2a$$
This is the slope of the tangent at any point $a$ on the curve, so the slope of the normal is $\frac{-1}{2a}$
So, the equation of the normal at any point $a$ on the curve,
$$y-a^2 = \frac{-1}{2a}(x-a)$$
As this normal passes through the point $(0,c)$, therefore
$$c-a^2=\frac{1}{2}$$
So, $$a^2 = \frac{2c-1}{2}$$
Therefore,
$$a = \pm \sqrt{\frac{2c-1}{2}}$$
Notice that this $a$ is any point at which the normal is drawn, and this value of $a$ is only defined when $c \geq \frac{1}{2}$
So, if $c > \frac{1}{2}$, there are two points possible $(+a\ and  \ -a)$, it means for any one value of $c$ which is greater than $\frac{1}{2}$, there are two normal lines possible. But, when $c = \frac{1}{2}, \ a = 0$, normal at $a = 0$ is actually the y-axis, which covers every $c$, and hence, $c > \frac{1}{2}$ . So, in total, there are total three normal lines possibles for $c > \frac{1}{2}$.
For $c \leq \frac{1}{2}$, $a$ is not defined for $c < \frac{1}{2}$, but for $c = \frac{1}{2}$, again, the normal is at $a=0$. So, for $c \leq \frac{1}{2}$, there is only one normal present.
