<s>[INST] Umbilical points of the ellipsoid. The following is Exercise (11) of Chapter 3 of Curves and Surfaces, 2nd edition, by Montiel and Ros:

Determine the umbilical points of the ellipsoid of equation
  $$
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
$$
  where $0 < a < b < c$.

My question is:

Why the umbilical points must have $y = 0$? Why can't it happen that $x = 0$ or $z = 0$?

The book provides a solution, that I worked out as follows (without quite understanding the conclusion):
Denote the ellipsoid by $S$. We begin by noting that $S$ is the inverse image of a regular value of the function
$$
f(x, y, z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}.
$$
Indeed, if $f(x, y, z) = 1$ then
$$
\nabla f(x, y, z) = 2 \left (\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right) \neq 0.
$$
It is clear that $S = f^{-1}(\{1\})$.
The gradient gives us a Gauss map defined on $S$:
$$
N(x, y, z) = \frac{\nabla f(x, y, z)}{|\nabla f (x, y, z)|}.
$$
Then, there exists a nonvanishing differentiable function $h$ defined on $S$ such that 
$$
h(x, y, z) N(x, y, z) = \left(\frac{x}{a^2}, \frac{y}{b^2}, \frac{z}{c^2}\right).
$$
Using the product rule and the fact that the right-hand side is linear, for all $p = (x, y, z) \in S$ and $v = (v_1, v_2, v_3) \in T_pS$ we have that 
$$
(dh)_p(v)N(p) + h(p)(dN)_p(v) = \left(\frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right).
$$
The point $p$ is umbilical if and only if $(dN)_p$ is a multiple of the identity. We claim that this happens if and only if the left-hand side vanishes after scalar multiplication by $N(p) \wedge v$. It is clear that the first term vanishes, since $(N(p) \wedge v) \perp N(p)$. Suppose $(dN)_p$ is a multiple of the identity. Then $h(p)(dN)_p(v) \in \langle v \rangle$ and the second term vanishes as well. On the other hand, assume that 
$$
\left(h(p) (dN)_p(v)\right) \cdot (N(p) \wedge v) = 0.
$$ 
Then $(dN)_p(v)$ lies in the plane spanned by $N(p)$ and $v$. But since $(dN)_p(v) \in T_pS$, if follows that it is a multiple of $v$, and therefore the claim follows. 
From the previous paragraph, we conclude that
$$
\left( \frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right) \cdot (N(p) \wedge v) = 0, 
$$
that is, $\left(\frac{v_1}{a^2}, \frac{v_2}{b^2}, \frac{v_3}{c^2}\right)$ lies on the plane spanned by $N(p)$ and $v$. Since these three vectors are linearly dependent, we have that 
$$
\begin{vmatrix}
\displaystyle \frac{v_1}{a^2} & \displaystyle \frac{v_2}{b^2} & \displaystyle \frac{v_3}{c^2} \\
v_1    & v_2     & v_3    \\
\displaystyle \frac{x}{a^2} & \displaystyle \frac{y}{b^2} & \displaystyle \frac{z}{c^2}
\end{vmatrix}
= 0, \quad \text{ if } \quad N(p)\cdot v = 0.
$$
But the determinant is a quadratic form in the variables $v_i$ which vanishes on a plane:
$$
q(v) = \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right)v_1 v_2 + \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right)v_2 v_3 + \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right)v_1 v_3
$$
Hence, the matrix of this quadratic form is
$$
A = \frac12 \begin{bmatrix}
0 & \displaystyle \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right) & 
 \displaystyle \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right)\\
\displaystyle \frac{z}{c^2}\left(\frac{1}{a^2} - \frac{1}{b^2}\right) & 0 & 
\displaystyle \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right)\\
\displaystyle \frac{y}{b^2} \left(\frac{1}{c^2} - \frac{1}{a^2}\right) & \displaystyle \frac{x}{a^2} \left(\frac{1}{b^2}- \frac{1}{c^2}\right) & 0
\end{bmatrix}.
$$
So far, so good. Now, the book argues that since this quadratic form vanishes on a plane, the determinant of $A$ must be zero.

Why is this so?

Moving forward, $\det A = 0$ if $x =0$ or $y = 0$ or $z = 0$. The book claims that it can only be $y = 0$. Why?
I have found this similar question, but I did not quite understand what is going on:
Umbilical points of Ellipsoid alternate method
Thanks in advance and kind regards
 [/INST] I think I got this.
Suppose $x = 0$. From $N(p) \cdot v = 0$ we have that 
$$
v_3 = - \frac{y}{b^2} \frac{c^2}{z}v_2.
$$
Then
$$
q(v) = \left( \frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) - \frac{y}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \right) v_1 v_2 = 0 \quad \forall v_1, v_2 \in \Bbb{R}.
$$
Then 
$$
\frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) = \frac{y^2}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \implies \frac{z^2}{c^2} = \frac{y^2}{b^2} \left( \frac{a^2 - c^2}{b^2 - a^2} \right) < 0,
$$
which is absurd.
If $y = 0$, by the same argument we arrive at 
$$
\frac{x^2}{a^2} = \frac{z^2}{c^2} \left( \frac{b^2 - a^2}{c^2-b^2} \right),
$$
which is feasible.
If in turn $z = 0$, then 
$$
\frac{x^2}{a^2} = \frac{t^2}{b^2} \left( \frac{a^2 - c^2}{c^2-b^2} \right) < 0, 
$$
which is also absurd. Hence, the only possible option is $y = 0$.
Any comments will be appreciated.
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