Problem
Find the umbilical points of the ellipsoid \( x^{2}/a^{2}+y^{2}/b^{2}+z^{2}/c^{2}=1 \) .
The ellipsoid in may be parametrized as
\[
\begin{align}
p(u,v)=(a\cos u\cos v,b\cos u\sin v,c\sin u)
\end{align}
\]
Solution
Partial differentiating with respect to
and
\[
\begin{align}
p_u&=(-a\sin u\cos v, -b\sin u\sin v, c\cos u)\\
p_v&=(-a\cos u\sin v, b\cos u\cos v, 0)\\
p_{uu}&=(-a\cos u\cos v, -b\cos u\sin v, -c\sin u)\\
p_{uv}&=(a\sin u\sin v, -b\sin u\cos v, 0)\\
p_{vv}&=(-a\cos u\cos v, -b\cos u\sin v, 0)
\end{align}
\]
The coefficients of the first fundamental form \(E\), \(F\), \(G\) may be found by
\[
\begin{align}
E&=p_u\cdot p_u\\
&=a^{2}\sin^{2}u\cos^{2}v+b^{2}\sin^{2}u\sin^{2}v+c^{2}\cos^{2}u\\
F&=p_u\cdot p_v\\
&=\left(a^{2}-b^{2}\right)\sin u\cos u\sin v\cos v\\
G&=p_v\cdot p_v\\
&=a^{2}\cos^{2}u\sin^{2}v+b^{2}\cos^{2}u\cos^{2}v
\end{align}
\]
We obtain the normal vector \( \nu \)
\[
\begin{align}
p_u\times p_v&=\cos u(-bc\cos u\cos v, -ac\cos u\sin v, -ab\sin u)\\
|p_u\times p_v|&=\cos u\sqrt{a^{2}b^{2}\sin^{2}u+b^{2}c^{2}\cos^{2}u\cos^{2}v+c^{2}a^{2}\cos^{2}u\sin^{2}v}\\
\nu&=\frac{p_u\times p_v}{|p_u\times p_v|}\\
&=\frac{1}{\Delta}(-bc\cos u\cos v, -ac\cos u\sin v, -ab\sin u)
\end{align}
\]
where \( \Delta=\sqrt{a^{2}b^{2}\sin^{2}u+b^{2}c^{2}\cos^{2}u\cos^{2}v+c^{2}a^{2}\cos^{2}u\sin^{2}v}\).
The coefficients of the second fundamental form \(L\), \(M\), \(N\) may be found by
\[
\begin{align}
L&=p_{uu}\cdot\nu\\
&=\frac{1}{\Delta}abc\\
M&=p_{uv}\cdot\nu\\
&=0\\
N&=p_{vv}\cdot\nu\\
&=\frac{1}{\Delta}abc\cos^{2}u
\end{align}
\]
We obtain Gaussian curvature \(K\) and mean curvature \(H\)
\[
\begin{align}
K&=\frac{LN-M^{2}}{EG-F^{2}}\\
&=\frac{1}{\Delta^{4}}a^{2}b^{2}c^{2}\\
H&=\frac{1}{2}\frac{EN-2FM+GL}{EG-F^{2}}\\
&=\frac{abc}{2\Delta^{3}}\left(a^{2}\left(\sin^{2}u\cos^{2}v+\sin^{2}v\right)+b^{2}\left(\sin^{2}u\sin^{2}v+\cos^{2}v\right)+c^{2}\cos^{2}u\right)
\end{align}
\]
Let
\[
\begin{align}
\textrm{(curvature)}=\frac{L\xi^{2}+2M\xi η+Nη^{2}}{E\xi^{2}+2F\xi η+Gη^{2}}=\lambda
\end{align}
\]
If principal curvature are \( \lambda_1 \), \( \lambda_2\) (\( \lambda_1\leq\lambda_2 \)), then
\[
\begin{align}
\textrm{(umbilical point)}&\Leftrightarrow \lambda = \lambda_1 = \lambda_2 = \sqrt{K} :\textrm{constant}\\
&\Leftrightarrow\frac{L\xi^{2}+2M\xi η+Nη^{2}}{E\xi^{2}+2F\xi η+Gη^{2}}=\sqrt{K}\\
&\Leftrightarrow \exists k \ \textrm{s.t.}\ L=\sqrt{k}E, M=\sqrt{k}F, N=\sqrt{k}G
\end{align}
\]
\(F=0\) by \(M=0\). Thus \(\sin u\cos u\sin v\cos v=0\). Then
- \(\sin u=0\)
Not exist the umbilical points, since \(a>b>c>0\). \(\sin v=0\)
Since \(\sin v=0\), \(\cos v=1\). Then the ellipsoid is \[ \begin{equation} p(u,v)=(a\cos u, 0, c\sin u) \end{equation} \] By \(\textrm{(umbilical point)}\Leftrightarrow E/L=G/N\) \[ \begin{align} a^{2}\sin^{2}u+c^{2}\cos^{2}u&=b^{2}\\ a^{2}\sin^{2}u+c^{2}\left(1-\sin^{2}u\right)&=b^{2}\\ \sin^{2}u\left(a^{2}-c^{2}\right)+c^{2}&=b^{2}\\ \sin^{2}u\left(a^{2}-c^{2}\right)&=b^{2}-c^{2}\\ c^{2}\sin^{2}u&=c^{2}\frac{b^{2}-c^{2}}{a^{2}-c^{2}}\\ c\sin u&=\pm c\sqrt{\frac{b^{2}-c^{2}}{a^{2}-c^{2}}}\\ z&=\pm c\sqrt{\frac{b^{2}-c^{2}}{a^{2}-c^{2}}} \end{align} \] Similarly, \[ \begin{align} a^{2}\left(1-\cos^{2}u\right)+c^{2}\cos^{2}u&=b^{2}\\ x&=\pm a\sqrt{\frac{b^{2}-a^{2}}{c^{2}-a^{2}}} \end{align} \] Therefore, the umbilical points of the ellipsoid on \(\mathbb{R}^3\) are \[ \begin{equation} \left(\pm a\sqrt{\frac{b^{2}-a^{2}}{c^{2}-a^{2}}},\,0,\,\pm c\sqrt{\frac{b^{2}-c^{2}}{a^{2}-c^{2}}}\right)\ \textrm{(Any double sign)} \end{equation} \]\(\cos u=0\)
Not exist the umbilical points, since \(a>b>c>0\).\(\cos v=0\)
Not exist the umbilical points, since \(a>b>c>0\).
Conclusion
The umbilical points of the ellipsoid on \(\mathbb{R}^3\) are the four points
\[ \begin{equation} \left(\pm a\sqrt{\frac{b^{2}-a^{2}}{c^{2}-a^{2}}},\,0,\,\pm c\sqrt{\frac{b^{2}-c^{2}}{a^{2}-c^{2}}}\right)\ \textrm{(Any double sign)} \end{equation} \]
the_umbilical_point_of_the_ellipsoid.pdf - Google ドライブ
はてなブログMathJaxへの（？）文句
\etaはηって書かなきゃいけないし，\textrm{const}が何故か弾かれるのでクソ．
と思ったのだけれどググる限り再現状況が良く分からないので結局何が悪いのか分からない．
