Finding the number of normals to a parabola 
Find the number of normals to the parabola $y^2=8x$ through (2,1)

$$$$
I tried as follows:
Any normal to the parabola will be of the form $$y=mx-am^3-2am$$
Since the point (2,1) lies on the normal, it satisfies the equation of the normal. Thus $$2m^3+2m+1=0$$
The number of real roots of $m$ in the above equation would indicate the number of normals which satisfy the condition given in the question. However, I got stuck while trying to manually calculate the roots.
I would be grateful if somebody could please show me a more efficient method of solving the question. Many thanks in advance!
 A: You have a cubic equation in $m$. You can find how many roots that equation has by looking at the discriminant
$$\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$
for the equation $ax^3+bx^2+cx+d=0$. In your case, $a=2,\ b=0,\ c=2,\ d=1$. So your discriminant is
$$\Delta=0-0+0-4\cdot 2\cdot 2^3-17\cdot 2^2\cdot 1^2=-172$$
This is negative, so your equation has exactly one root. Your problem does not require actually finding the solution, so you are done (if, of course, your other work is correct).
For confirmation, you can look at the derivative of your equation:
$$\frac{dy}{dm}=6m^2+2$$
That derivative is always positive, so the function represented by $y$ is strictly increasing and there can be at most one zero. A cubic function must have at least one zero, so we know there is exactly one zero.
A: This is a really interesting question, and although a solution really does involve a discriminant, the general answer is so pleasing from an esthetic standpoint, and my comments above were so wrong, that I thought I’d show you a treatment of the problem.
I’ll work with the parabola $Y=X^2/2$, chosen for no particular reason. But since all parabolas are of the same shape, this will do for all. Note that its focus is at $(0,1/2)$.
Take a general point $(x,y)$ in the plane, and ask what the points $(\xi,\xi^2/2)$ are on the parabola with the property that the line from $(x,y)$ to $(\xi,\xi^2/2)$ is orthogonal to the curve. The tangent to the curve has slope $\xi$, so the orthogonal has slope $-1/\xi$, and we ask for the $\frac{\xi^2}2-y=-\frac1\xi(\xi-x)$, thus $\xi^3-2y\xi=2x-2\xi$, so that, given $x$ and $y$, we get a cubic for $\xi$:
$$
\xi^3-2(y-1)\xi-2x=0\,,
$$
and this is the equation whose roots $\xi$ we want to examine qualitatively. Since the discriminant of $X^3-aX+b$ is $4a^3-27b^2$, the discriminant of our polynomial is $32(y-1)^3-108x^2$, or in other words
we want to look at the curve $(Y-1)^3=\frac{27}8X^2$. This is a cusp cubic curve, note that the cusp is at $(0,1)$, twice as far from the vertex as the focus is.
The upshot? Below the curve $(Y-1)^3=\frac{27}8X^2$, there is only one normal to be drawn from $(x,y)$ to the parabola; above it, three; and on the curve, two normals to the parabola, except at the cusp, where all three coalesce, since the circle centered at $(0,1)$ of radius one has a contact with the parabola of order four.
A: It's easy.. If normals are drawn from a point $(h, k)$ to a parabola $y^2 = 4ax.$
$1).$ There are $3$ normals if $h>2a$ 
$2).$ There is only one normal if $h<2a$
