Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$
I think this is correct.
and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.
I don't think this is correct.
Let $(s,t)$ be a point on $x^2=4by$, i.e. $s^2=4bt$. Since $2x=4by'\Rightarrow y'=\frac{x}{2b}$, the equation of normal to the parabola at $(s,t)$ will be $y-t=-\frac{2b}{s}(x-s)\iff y=-\frac{2b}{s}x+2b+t$. Let $k=-\frac{2b}{s}$. Then, since $s=-\frac{2b}{k}$, the equation of the normal can be written as
$$y=kx+2b+\frac{s^2}{4b}=kx+2b+\frac{1}{4b}\left(-\frac{2b}{k}\right)^2=kx+2b+\frac{b}{k^2}.$$
So, in our case, we have
$$y=m_1x+2b+\frac{b}{m_1^2}\qquad\text{and}\qquad y=m_2x+2b+\frac{b}{m_2^2}.$$
Thus, since $y=m_1x+\frac{a}{m_1}$ and $y=m_1x+2b+\frac{b}{m_1^2}$ are the same lines,
$$\frac{a}{m_1}=2b+\frac{b}{m_1^2}\iff 2bm_1^2-am_1+b=0.$$
Now, the discriminant is positive, so $(-a)^2-4\cdot 2b\cdot b\gt 0$, i.e. $a^2\gt 8b^2$.
As harry pointed out in the comments, one has to consider the case where one of the tangents is the $y$-axis.
Two tangents drawn to the parabola $y^2=4ax$ from $(0,k)$ where $k\not=0$ are $x=0$ and $y=\frac akx+k$. It follows from $u^2=4bv,\frac ak=-\frac{2a}{u}$ and $k=2b+v$ that $bk^2-a^2k+2a^2b=0$. So, one has to have $a^2\geqslant 8b^2$. If $a^2=8b^2$, then two tangents drawn to the parabola $y^2=4ax$ from a point $(0,4b)$ are $x=0$ and $ax-4by+16b^2=0$ (at $(2a,8b)$) which are normals to the parabola $x^2=4by$ (at $(-a,2b)$).
In conclusion, the answer is $a^2\geqslant 8b^2$.
Corresponding to parameter value $t$, the gradient of the tangent is $\frac 1t$.
Therefore, from the angle formula, you have $$\left|\frac{\frac{1}{t_2}-\frac{1}{t_1}}{1+\frac{1}{t_1t_2}}\right|=\left|\frac{t_1-t_2}{1+t_1t_2}\right|=1$$
You also have $\frac xa=t_1t_2$ and $\frac ya=t_1+t_2$
Writing the first equation as $$|t_1-t_2|=|1+\frac xa|$$ and squaring both sides, we get
$$\left(1+\frac xa\right)^2=(t_1-t_2)^2=(t_1+t_2)^2-4t_1t_2=\left(\frac ya\right)^2-4\frac xa$$
Hence the result follows immediately
Best Answer
You have almost given the general condition yourself.
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/