Theorem: 4 normals can be drawn from any point to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. $( a > b > 0)$
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.