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.

• What is a cone surface? – user7530 Sep 5 '15 at 4:43

Surfaces of zero Gauss curvature $K$ are the following:

1. All cones
2. All cylinders as special case of cones
3. Developable helicoids,and,
4. 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.

• I think that at least case 2 is mostly excluded: usual cylinder doesn't have the property that all tangent planes pass through one common point. – Evgeny Sep 5 '15 at 12:28
• In my textbook, there is a theorem which says that "Surface of zero Gauss curvature and without umbilic points is developable". In my opinion, if we drop the condition "without umbilic points", the conclusion may be wrong. – sigma_k Sep 5 '15 at 13:52
• Unable to see difference between cone and cylinder in this respect. ( Salutations to the city of Lyapunev and Lobachevsky.) – Narasimham Sep 5 '15 at 19:40

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 $a_0$.

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 $\mathbb{R}^3$