Umbilical points of Ellipsoid alternate method I'm having serious trouble finding the umbilical points of the ellipsoid represented by
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,    \;\;\;a,b,c\neq 0.$$
My first thought was to use the parametrization
$$\mathbf{x}(u,v)=(a\sin(u)\cos(v),b\sin(u)\sin(v),c\cos(u)),$$
for $0<u<\pi$ and $0<v<2\pi$, compute the first and second fundamental forms, etc., but this is a nightmare. After doing some researching (and on the back solutions of Do Carmo) I came across I suppose what would be an alternate method which doesn't dig directly into a parametrization. It is explained slightly at the end of the pdf:
http://www.math.umn.edu/~voronov/5378/sample1.pdf
which essentially states to notice that $N_1=(\frac{x^2}{a^2},\frac{y^2}{b^2},\frac{z^2}{b^2})$ (the gradient) is such that $N_1=fN$, for some $f$ such that $|f|=|N_1|$, where $N$ is the unit normal vector to surface, as well as notice a point on a curve $\alpha(t)=(x(t),y(t),z(t))$ lying on the ellipsoid is an umbilical point iff the vector triple product
\begin{equation}
\left(\frac{dN_1}{dt}\wedge \alpha '\right)\cdot N_1=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)
\end{equation}
which I mostly understand. Then it says use some trickery by multiplying a $\frac{z}{c^2}$ to the equation and put it in terms of $x,x',y',$ and $y$ like this...
The way I understand it you should start with equation (1) in this form
$$
\left(\left(\frac{x'}{a^2},\frac{y'}{b^2},\frac{z'}{c^2}\right)\wedge\left(x',y',z'\right)\right)\cdot\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right)=0.
$$
Then, making things more complicated, (plugging everything in and doing the computation) we have
$$
\frac{xy'z}{a^2b^2}-\frac{xy'z'}{a^2c^2}+\frac{x'yz'}{b^2c^2}-\frac{x'yz'}{a^2b^2}+\frac{x'y'z}{a^2c^2}-\frac{x'y'z}{b^2c^2}=0.
$$
Multiplying $\frac{z}{c^2}$ to both sides gives
$$
\frac{xy'z'z}{a^2b^2c^2}-\frac{xy'z'z}{a^2c^4}+\frac{x'yz'z}{b^2c^4}-\frac{x'yz'z}{a^2b^2c^2}+\frac{x'y'z^2}{a^2c^4}-\frac{x'y'z^2}{b^2c^4}=0
$$
From here I suppose one would use the original equation for the ellipsoid as well as implicit derivative, $\frac{2zz'}{c^2}=-\frac{2yy'}{b^2}-\frac{2xx'}{a^2}$ to get rid of $z$ and $z'$. However, when I do that it starts getting pretty messy and I'm starting to believe I'm not quite understanding the method correctly. I also believe $y=0$ should satisfy this equation, but that's not quite working out, which also leads me to believe that I'm wrong in my thought.
Any opinions/suggestions would be greatly appreciated. Thank you
 A: An umbilical point $x$ of a surface $S$ is characterized by $dN_x(v)=kv$ for every $v$ in $T_xS$.
If $v=dx/dt$ then $dN/dt=kdx/dt$ is equivalent to $\det(dN/dt,N,dx/dt)=0$. 
If $N_1=fN$ then $N_1'=f'N+fN'$ and $\det(N_1',N_1,x')=f^2\det(N',N,x')$.
Then taking $N_1=(x_i/a_i)$, $N'=(v_i/a_i)$ and $x'=(v_i)$ for the ellipsoid $E$:
$$\sum_{ i=1}^3 x_i^2/a_i=1 \quad \text{with} \quad 0 < a_1 < a_2 < a_3$$ we obtain that $$\det(N_1',N_1,x')v_3/a_3=[(a_3-a_2)x_1v_2+(a_3-a_1)x_2v_1](-x_3v_3/a_3)+[v_1v_2(a_2-a_1)]x_3^2/a_3.$$ Substituting $-x_3v_3/a_3~$ by $~x_1v_1/a_1+x_2v_2/a_2$ we deduce that the umbilical points $(x_i)$ of $E$ satisfy that
$$v_1^2x_1x_2(a_3-a_1)/a_1+v_2^2x_1x_2/a_2+v_1v_2\Delta=0$$ for every $(v_1,v_2)$, i.e. $x_1x_2=0$ and $$\Delta:=-x_1^2(a_3-a_2)/a_1+x_2^2(a_3-a_1)/a_2+x_3^2(a_2-a_1)/a_3=0.$$
If $0<a_1<a_2<a_3$ then the only solutions are given by $$x_2=0~, \quad x_1^2(a_3-a_2)/a_1=x_3^2(a_2-a_1)/a_3$$ and $$x_1^2/a_1+x_3^2/a_3=1~,$$ that is $$x_1=\pm\sqrt{a_1(a_2-a_1)/(a_3-a_1)}~, \quad x_2=0~, \quad \text{and} \quad x_3=\pm\sqrt{a_3(a_3-a_2)/(a_3-a_1)}$$
