Umbilics on the ellipsoid Show that, if p, q and r are distinct positive numbers, there are
exactly four umbilics on the ellipsoid $$\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1$$ 
What happens if $p$, $q$ and $r$ are not distinct? 
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To show that I considered the parametrization $\sigma (u,v)=(p\cos u\sin v, q\sin u\sin v, r\cos v)$. 
Then the principal curvatures are the roots of 
$\begin{vmatrix} 
L-\kappa E & M-\kappa F\\ 
M-\kappa F & N-\kappa G 
\end{vmatrix}=0$. 
I found the following: 
$$\kappa^2 \sin^2 v \\ \left [((p^4+q^4)\sin^2 u\cos^2 u+p^2q^2(\cos^4 u+\sin^4 u))\cos^2 v+r^2\sin^2{v}(p^2\sin^2 u+q^2\cos^2 u)-(q^2-p^2)\sin^2u \cos^2u \cos^2v\right ]-\kappa \frac{pqr\sin^2 v}{\sqrt{\sin^2 v(q^2r^2\cos^2 u+p^2r^2\sin^2 u)+p^2q^2\cos^2 v}} \left [(p^2\cos^2 u+q^2\sin^2 u)(\cos^2 v+1)+r^2\sin^2 v)\right ]+\frac{p^2q^2r^2\sin^2 v}{\sin^2 v(q^2r^2\cos^2 u+p^2r^2\sin^2 u)+p^2q^2\cos^2 v}=0 $$ 
Can this be correct? Can we simplify it? 
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 A: Consider a parametrization $f$ :
$$ x=a\cos\ t \cos\ s,\ y=b\cos\ t\sin\ s,\ z=c\sin\ t $$
Define $$ A:=\cos\ t,\ B=\sin\ t,\ C=\cos\ s,\ D=\sin\ s $$
 Then $$
f_t=(-aBC,-b BD, c A ) ,$$ $$ f_s=(-aAD, b AC, 0 )
$$
$$ f_{tt} =(-aAC,-b AD,-cB) $$
$$ f_{st} = (aB D, -bBC,0) $$
$$ f_{ss} = (-aAC,-bA D,0 ) $$
Hence we have first fundamental forms : $$ E=a^2 B^2 C^2 +
b^2 B^2 D^2 + c^2 A^2
$$
$$ F=(a^2-b^2) ABCD $$
$$ G=A^2 ( a^2 D^2 + b^2C^2)  $$
so that $$ EG-F^2 = A^2\{ a^2b^2 B^2D^2 + a^2c^2 A^2D^2 + b^2 c^2
A^2 C^2 \}
$$
Then unit normal is
$$ N = \frac{f_t\times f_s}{|f_t\times f_s|}  =
(-bcA^2 C, ac A^2 D, -ab  AB ) l $$ where $$ l= \frac{1}{ |A| \sqrt{
(bc AC)^2 + (ac AD)^2 + (abB)^2 } } $$ And second
fundamental forms are
$$ e= (N,f_{tt})= abc\ A\{ A^2 C^2 - A^2 D^2 +B^2 \} l $$
$$ f=(N, f_{st})= -2abc\ A^2BCD l  $$
$$ g= (N, f_{ss})= abc \ A^3(C^2-D^2) l  $$
Then
$$ \frac{ eg-f^2 }{(abcA^2l)^2} = A^2 ( C^2-D^2)^2  +
4 B^2C^2D^2 + B^2 (C^2-D^2) :=m  $$
And $$ \frac{1}{2} \frac{gE -fF+ eG }{abc\ A^3l} =  \frac{1}{2} \{ (
a^2 D^2 + b^2C^2) \{ A^2 C^2 - A^2 D^2 +B^2 \}  $$ $$+2(a^2-b^2)
(BCD)^2 +( a^2 B^2 C^2 + b^2 B^2 D^2 + c^2 A^2)
 (C^2-D^2) \}   $$
$$ =\frac{1}{2}\bigg[ a^2 \{  C^2 D^2 - A^2D^4+B^2C^4 + B^2D^2 \} $$ $$+ b^2 \{ -C^2D^2 +
A^2C^4 + B^2C^2-B^2 D^4 \} + c^2A^2 (C^2-D^2) \bigg] :=n$$
By $H^2=K$, we have $$ \bigg( \frac{n \ abc\ A^3l}{EG-F^2 } \bigg)^2
= \frac{m\ (abc\ A^2l)^2}{EG-F^2} \Rightarrow n^2A^2 =m(EG-F^2) $$
$$\frac{1}{A^2} m(EG-F^2) -n^2$$
$$= ( a^2b^2 B^2D^2 + a^2c^2 A^2D^2 + b^2 c^2 A^2 C^2 )( A^2
(C^2-D^2)^2 + 4 B^2C^2D^2 + B^2(C^2-D^2)  )$$ $$ - \frac{1}{4} ( a^2
( C^2 D^2 -A^2D^4+B^2C^4 + B^2D^2 ) $$ $$+ b^2 ( -C^2D^2 +A^2C^4 +
B^2C^2-B^2 D^4 )+c^2A^2 (C^2-D^2) )^2
$$
$a=b,\ C=0,\ D=1$ Case : We will consider $z\geq
0$ Then $(0,0,c)$ is umbilic. Assume that $0< A\leq 1$ Then
$$\frac{1}{A^2} m(EG-F^2) -n^2 = -\frac{1}{4} \{
a^2(-3A^2+2) + A^2c^2 \}^2$$
Hence $$ c= \frac{a\sqrt{3A^2-2}}{A} $$
