Surface with all tangent plane pass through one common point is a cone surface I want to prove that a surface with all tangent plane pass through one common point is a cone surface (a surface like this $\mathbf{r}(u,v)=\mathbf{a}_0+v\mathbf{b}(u)$, here $\mathbf{a}_0$ is a constant vector). 
Hints1: use natural coodintates; Hints2: use the orthonormal frame and exterior differential forms. 
For both methods, up to now, I can only prove that the Gauss curvature of the surface is zero. I do not know how to proceed? Any ideas or comments will be helpful.
 A: Surfaces of zero Gauss curvature $K$  are the following:


*

*All cones

*All cylinders as special case of cones

*Developable helicoids,and,

*All developable surfaces that are isometric to them.


If you proved $K =0$ then you already proved it belongs to one among them, the cone is included. 
You cannot prove that it is a conical surface. Any one of them can pass to to the other by bending or isometric mapping.
In each case you need to show where  $r,a,b$ vectors are.
Cases $1$ to $3$ have generators straight, case $4$ has the line of regression straight.
A: Suppose the common point is $a_0$. Let $p$ be a point on your surface and consider the curve $\gamma(s)$ satisfying $\gamma'(s) = a_0 - \gamma(s)$ and $\gamma(0) = p$. Assuming sufficient regularity of your surface this curve is tangent to the surface. Moreover for any vector $v$ orthogonal to $a_0-p$ we have that
$$\gamma'(s)\cdot v = \frac{d}{ds}(\gamma(s)\cdot v) = a_0 \cdot v - \gamma(s)\cdot v$$
which for initial condition $\gamma(0)\cdot v = a_0 \cdot v$ has unique solution $\gamma(s)\cdot v = a_0\cdot v$ and $\gamma'(s)\cdot v =0$.
So $\gamma$ is a ruling of your surface and all such rulings, when extended, intersect at the common point $a_0$.
A: For a surface $S$ we have a point $p$ in the assumption Note that
for $x\in S$, we have nonvanishing unit vector field
$e_1:=\frac{\overrightarrow{xp}}{|\overrightarrow{xp}|}$ Note that
this vector field has an extension $V$ on $\mathbb{R}^3$ which is
$\frac{\overrightarrow{yp}}{|\overrightarrow{yp}|}$ at $y$ That is
$V|S=e_1$ Note that flow of $V$ is a line. So flow in $S$ is line in
$\mathbb{R}^3$
