Surface with all tangent plane pass through one common point is a cone surface

Question

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.

Answer 1

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.

Answer 2

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 $a_0$.