Condition for 3 distinct normals from a point $(h,k)$ to the parabola $y^2=4ax$

Question

What are the conditions in terms of $h,k$ to be able to draw three distinct normals to the parabola $y^2= 4ax$ ?

Normal to the parabola from $(h,k)$ is given by: $am^3 +(2a-h)m+k=0$. This equation can yield three distinct slopes $m_1,m_2,m_3$ if $\Delta>0$ (source).

The $\Delta$(discriminant) of the equation is given by $-4a (2a-h)^3-27a^2k^2$.

For it to be greater than $0$ when $a>0$, I derived $h>2a$ and $k<-2\sqrt{2}a$. How do I derive the other constraints on $h,k$ i.e. $k>4\sqrt{2}{a}$ and $h>8a$ ?

Answer 1

1. Apply the condition for normal $y=mx+am^3-2am$ for the given parabola.
2. Let $m_1$, $m_2$ and $m_3$ be the slopes of the required normals to the parabola.
3. For three distinct normals the slopes should satisfy the following condition: $m_1+m_2+m_3=0$.

Answer 2

You have almost given the general condition yourself.

The Δ(discriminant) of the equation is given by $-4a (2a-h)^3-27a^2k^2$.

If discriminant of the cubic equation is positive, then the cubic equation has 3 real roots.

The discriminant of the cubic normal equation must be positive to have 3 real and distinct slopes, that means, three distinct normals from that particular point (h, k).

So the required condition for three normals to be drawn from $(h,k)$ to the parabola $y^2 =4ax$ is just

$$ -4a (2a-h)^3-27a^2k^2 > 0 $$

On rearranging, we get a compact representation :

$$ \biggl(\frac{h-2a}{3}\biggr)^3> a\biggl(\frac{k}{2}\biggr)^2 $$

You need to know one of the coordinates $h$ or $k$, or conditions on one of the coordinate to obtain constraint for the other coordinate.

I have referred this site: https://brilliant.org/wiki/cubic-discriminant/