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

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 $h>8a$ ?

• math.stackexchange.com/questions/2133724/… – Fawad Nov 23 '17 at 17:13
• "when $a\gt 0$, I derived $h\gt 2a$ and $k\lt -2\sqrt 2a$" This is incorrect. For example, $(a,h,k)=(1,3,1/3)$ satisfies $\Delta\gt 0,a\gt 0$ and $h\gt 2a$, but does not satisfy $k\lt -2\sqrt 2a$. – mathlove Nov 24 '17 at 6:20