In Elementary Differential Geometry by A Pressley there is a problem asking to prove if $p,q,r$ are distinct positive real numbers then the ellipsoid In $\frac{x^{2}}{p^{2}}+\frac{y^{2}}{q^{2}}+\frac{z^{2}}{r^{2}}=1 $ has exactly 4 umbilics. I need some clarity regarding the solution given by the author.
He starts by assuming $z \neq 0$ and then show that $x,y$ both cannot be non zero. Later he argues if $x=0$ then $y = \pm q \sqrt{ \frac{q^{2}-p^{2}}{q^{2}-r^{2}}}$ and $z = \pm r \sqrt{ \frac{r^{2}-p^{2}}{r^{2}-q^{2}}}$.
It is clear that these equations satisfy $\frac{y^{2}}{q^{2}}+ \frac{z^{2}}{r^{2}}=1$. But I am not getting how this answer is obtained ? Does it have any thing to do with first and second fundamental forms ? ( Here it will be $E=F=0=L=M$)
At some point I have typed a detailed solution to that problem. Maybe it helps.
We compute the umbilical points of the ellipsoid $E=\left \{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \right \}$. Let $$ f(x,y,z) = \left(\frac{x}{a^2} + \frac{y}{b^2} + \frac{z}{c^2}\right)^{-1/2} $$ Then a unit normal vector field to $E$ is given by $N(x,y,z) = f(x,y,z) (\frac{x}{a^2} , \frac{y}{b^2} , \frac{z}{c^2})$ and for a tangent vector $v = (v_1,v_2,v_3)\in T_{(x,y,z)}E$ one computes the shape tensor as \begin{equation} \tag{0} \begin{array}{rcl} Av &=& DN[v] = \langle \text{grad}f \vert v \rangle \left(\frac{x}{a^2}, \frac{y}{b^2} , \frac{z}{c^2}\right ) + f \text{diag}(a^{-2},b^{-2},c^{-2}) v \\\\ &=& -f(x,y,z)^3 \left(\frac{xv_1}{a^4} + \frac{yv_2}{b^4} + \frac{zv_3}{c^4}\right) \left(\frac{x}{a^2} , \frac{y}{b^2} , \frac{z}{c^2}\right ) + f(x,y,z) \left(\frac{v_1}{a^2} , \frac{v_2}{b^2} , \frac{v_3}{c^2}\right ) \end{array} \end{equation} At a point $p=(x,y,z) \in E$ the tangent space is spanned by $$ w_1 = (0,-b^2z,c^2 y), \quad w_2 = (-a^2z,0,c^2x), \quad w_3 = (-a^2y,b^2x,0) $$ If $p$ is umbilical, the vector $Aw_1$ is proportional to $w_1$, in particular, the first component of $Aw_1$ is zero. In general we see that the $i$th component of $Aw_i$ is zero. Computing the respective components with $(0)$ and setting them to zero gives us three equations: $$ 0 = xyz \left( \frac{1}{b^2} - \frac{1}{c^2} \right) = xyz \left( \frac{1}{a^2} - \frac{1}{c^2} \right) = xyz \left( \frac{1}{a^2} - \frac{1}{b^2} \right) $$ If $E$ is a sphere, then all the brackets are zero, but in this case we already know that all points are umbilical. On the other hand if $E$ is not a sphere, then one of the brackets is nonzero which implies that $xyz=0$. We have shown:
Lemma If $E$ is a sphere, then all points are umbilical. If $E$ is not a sphere, then all umbilical points lie on a plane (through the origin) which is normal to one of the coordinate axes.
Now let's focus on the plane $\{y=0\}$ and try to find out whether a point $p = (x,0,z) \in E$ is umbilical. If $p$ is umbilical, we have $Av= \lambda v$ for some $\lambda \in \mathbb{R}$. In particular this is true for $w_1 = (0,-b^2z,0)$ and $ w_3 = (0,b^2x,0) $ which yields the following equations: $$ - \lambda b^2 z = - f z \quad \text{and} \quad \lambda b^2 x = f x $$ Note that not both $x$ and $z$ can be zero thus one of the above equations implies that \begin{equation} \tag{1} \lambda b^2 = f \end{equation} But we also have $Aw_2= \lambda w_2$. This yields two equations which both simplify (by using $(1)$) to the following relation: \begin{equation} \tag{2} \frac{x^2}{a^2} (b^2 - c^2) = \frac{z^2}{c^2} (a^2 -b^2) \end{equation} If we have $a = b$, then $x=0$ and thus $p = (0,0,\pm c)$. If $b = c$, then $z = 0$ and thus $p=(\pm a, 0, 0)$.\ In the generic case, namely if $a,b,c$ are all pairwise different, $(2)$ together with the condition $\frac{x^2}{a^2}+ \frac{z^2}{c^2} = 1$ implies $$ x^2 = a^2 \left( \frac{a^2 - b^2}{a^2-c^2} \right) \quad \text{ and } \quad z^2 = c^2 \left( \frac{b^2 - c^2}{a^2-c^2} \right) $$ Note that these equations only have real solutions if $a < b < c$ or $a > b > c$. Of course it was completely arbitrary that we concentrated on the plane $\{y=0\}$, but after relabeling if necessary we can conclude the following: If an umbilical point lies on the plane normal to a certain coordinate axis, then the length of the corresponding principal axis of $E$ has to lie between the lengths of the other two principal axes. We have proven one direction of the following lemma.
Lemma If the lengths of the principal axes satisfy $a \le b \le c$ (but not $a=b=c$), then all all umbilical points are given by $$ \left( \pm a \sqrt{\frac{a^2 - b^2}{a^2-c^2}} , 0 , \pm c \sqrt{\frac{b^2 - c^2}{a^2-c^2}} \right) $$ where any sign combination is allowed. In particular there are exactly four umbilical points in the generic case ($a<b<c$) and two umbilical points in the case that two principal axes have the same length ($a=b$ or $b=c$).
It is left to show that the named points are indeed umbilical. Instead of tracing back all conclusions and asking for equivalence it is probably easiest to just plug them into $(0)$ again.