One may approach the problem in a more algebraic way. If the arbitrary point within the ellipse is O(x_{0},y_{0}) and a point on the ellipse is given parametrically by P(acos(t),bsin(t)), then the desired geometric orthogonality condition is that the vector OP be perpendicular to the tangent vector T to the ellipse at P or
áacos(t)x_{0},bsin(t)y_{0}ñ·áasin(t),bcos(t)ñ =0 . 

(7) 
Simplifying this condition (7) and eliminating sin(t) from it gives the location of the normal points as solutions to the quartic equation in cos(t):
(a^{2}b^{2})^{2}cos^{4}(t) 2ax_{0}(a^{2}b^{2})cos^{3}(t) +(a^{2}x_{0}^{2}+b^{2}y_{0}^{2} (a^{2}b^{2})^{2})cos^{2}(t) 

+2ax_{0}(a^{2}b^{2})cos(t) a^{2}x_{0}^{2}=0. (8) 

This is equivalent to a quartic equation in x(t)=a cos(t), which in turn is the result of using the equation of the hyperbola of Apollonius to eliminate y from the pair of quadratic equations for the hyperbola and ellipse.
Older books on the theory of equations discuss the nature of the roots of third and fourth degree polynomials, e.g.[4,5]. They develop the theory of the discriminant of the cubic and quartic, generalizing what every student knows about quadratic equations, and which is conveniently programmed into modern computer algebra systems. In fact as in the quadratic case, the discriminant can be defined as a product of the squared differences of all the distinct pairs of roots of the polynomial, modulo a normalizing constant, so it is zero precisely when there are multiple roots of the polynomial and nonzero otherwise, while its sign is correlated with the number of real roots.
If q(x)=a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}, a_{4} ¹ 0 is an arbitrary quartic with real coefficients, then one may define its discriminant in terms of these coefficients, the vanishing of which implies the existence of a multiple root as in the quadratic case. The discriminant of q is given by the formula

D(a_{4},a_{3},a_{2},a_{1},a_{0}) 

=4 
æ
è 
a_{2}^{2}
3

+ 4a_{0} a_{4} a_{1}a_{3} 
ö
ø 
3






27 
æ
è 
a_{1}^{2}a_{4}  a_{0}a_{3}^{2}  
2a_{2}^{3}
27

+ 
a_{1}a_{2}a_{3}
3

+ 
8a_{0}a_{2} a_{4}
3

ö
ø 
2

(9). 


In our case from (8) we have
a_{4}=(a^{2}b^{2})^{2} a_{3} = 2ax_{0}(a^{2}b^{2})=a_{1}, 

a_{2}=a^{2}x_{0}^{2}+b^{2}y_{0}^{2}(a^{2}b^{2})^{2} a_{0} = a^{2}x_{0}^{2}. (10) 

Substituting (10) into (9) and imposing the condition (5) that (x_{0},y_{0}) is on the evolute, one may show that D = 0 for all 0 £ t < 2p. Thus for these points there are either three normals from the point or two when the point is a cusp. One can also show that except for the discriminant vanishing on the axes, where the cosine must have at least a repeated root by symmetry, one has D < 0 for points outside the evolute, implying two distinct real roots and hence two normals, and D > 0 for points within the evolute, implying either four distinct real roots (and hence four normals) or two distinct pair of complex conjugage roots. However, the latter would imply that there are no normals, but as noted in the introduction there are always at least two so this cannot happen.
The reader interested in seeing more details of this development and a chance to investigate this problem or similar ones may download a Maple worksheet at
http://www3.villanova.edu/maple/misc/frenetellipse.htm or simply view the worksheet in web page format. This is a beautiful example of how, empowered by a computer algebra system, one can follow one's nose in uncovering elegant mathematical structure that would otherwise be unreachable in practice. It is also a useful lesson on the increasing importance of the use of computer algebra systems in doing mathematics.