4 normals can be drawn from any point to an ellipse proof doubt: How do we know the key quartic has 4 distinct real roots? Theorem: 4 normals can be drawn from any point to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. $( a &gt; b &gt; 0)$

I had a doubt in the proof of the above statement given in my book.

Proof given in Book.
The normal at the point with eccentric angle $\phi$ is $ax\sec\phi - by \csc\phi = a^2 - b^2 $. Let P$(h,k)$ be any point in $\mathbb{R^2}$. If the normal passes through P: 
$$ah\sec\phi - bk \csc\phi = a^2 - b^2  = a^2e^2 $$
Let $ t = \tan \frac{\phi}{2}$, then $\cos \phi = \frac{1 - t^2}{1 + t^2}$ and $\sin \phi = \frac{2t}{1 + t^2}$. Substituting and rearranging, gives:


  $$bkt^4 + 2t^3(ah + a^2e^2) + 2t(ah - a^2e^2) - bk = 0 $$


This is a quartic in $t$, hence has 4 roots, corresponding to 4 points on the ellipse which are conormal. 


  Doubt: My question is how do we know that the roots of that quartic equation are real and distinct? Because if not, then the theorem is not yet proved. I looked up the discriminant of a quartic on Wikipedia (https://en.wikipedia.org/wiki/Quartic_function), but the inequalities are too difficult to handle. How can I show that the roots are all real and distinct?
I think the theorem is simply false. Here is a rendering of the set of lines normal to an ellipse $(a=1, b=1/2)$:



There is a region (if I had to guess, bounded by the evolute of the ellipse: it would be interesting to prove this) where every point is indeed intersected by four normals to the ellipse. Outside this region, only two normals can be drawn.
Here is an example of such 4 normal:



Hope this helps!