What is the relationship between line of curvature and umbilical point? I am guessing whether or not the following statement is true:
All the points lie on a line of curvature of a connected curve are umbilical points. Conversely, given an umbilical point on a surface, it must lie on some line of curvature.
Is my guess correct? Any brief hints for the proof? If it is wrong, how could we explain it?
I could not find any theorem that relates umbilical points with line of curvature. Does any such theorem exist?
Thanks for the help
 A: In differential geometry there are lots of curvatures. In your case normal curvature is what you are looking at.
Compound Gauss curvature surfaces have either positive or negative sign. The sign allows you to divide either type (elliptic/hyperbolic points) into 4 quadrants, the dividing lines having of max/min or extreme curvatures. 
Cyclic variation of normal curvature takes place by Euler's law of curvature with respect to normal plane of intersection rotation $\psi$ to either of two principal planes.
$$ k_n = k_1 \cos^2 \psi + k_2 \sin ^2 \psi $$
When normal curvature is same irrespective of normal section such demarcating lines cannot be drawn.This is the case of umbilical points for all $\psi$
$$ k_1 = k_2 =k $$
Your understanding may need to be brought in line with the above.
All points on a line of curvature are umbilical.. this is incorrect. 
The sharp end points of a prolate ellipsoid of revolution are umbilical points.
In the example given by Weatherburn, Differential Geometry, pp 89, Ex 9, Chapter IV-40  we see for ellipsoid surface as also on page 82,  DJ Struik  2nd Dover Edition :
$$ \frac {x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$
whose lines of curvature interestingly go around four umbilical points.
https://en.wikipedia.org/wiki/Umbilical_point
EDIT1:
Wiki above also gives the same  plot. The umbilical point is enclosed between a set of two opposite signs curvature. plots.
