Prove that a surface is a conical surface The question follows that

If the tangent plane of a surface passes a fixed point, then it is a conical surface.(i.e. it can be parametrically represented as $\vec{r}(u,v)=\vec{a}+v\vec{b}(u)$. where $\vec{a}$ is constant vector).

I have the following attempt:
It is known that $\langle \vec{r}-\vec{a},\vec{n}\rangle=0$ by the assumption, where $\vec{a}$ is constant vector, $\vec{n}$ is the normal vector. Sowe get
$$\langle \vec{r}-\vec{a},\vec{n_u}\rangle=\langle \vec{r}-\vec{a},\vec{n_v}\rangle=0.$$
Then one of $\vec{n_u},\vec{n_v}$ must be equal to zero. WLOG $\vec{n_v}=0$,we may assume $\vec{n}=\vec{f}(u)$. Also, this imples $\vec{r}_v=\alpha(u,v)(\vec{r}-\vec{a})$ since $\vec{r}_v $ is perpendicular to both $\vec{n}$ and $\vec{n}_u$.
But how to proceed this proof? I have been thinking if this for a long time. Thanks for you helping hand.
 A: Here's a sketch for you to fill in:
Since $\vec n$, $\vec n_u$, and $\vec n_v$ are linearly dependent and the latter two are orthogonal to the first, we conclude that $\vec n_u$ and $\vec n_v$ are linearly dependent. In particular, the Gauss map has rank at most $1$. Assuming it is $1$, choose a parametrization so that the $n_v=0$. It now follows from the Mainardi-Codazzi equations that the $v$-curves are lines, so that the surface is a ruled surface. Last, parametrizing the surface as a ruled surface $x(u,v) = \alpha(u) + v\beta(u)$, it follows that $\alpha'$, $\beta$, and $\beta'$ are everywhere linearly dependent. Now you should be able to use the fact that all the tangent planes have a common point to show that $\alpha + \lambda\beta=\text{constant}$ for some function $\lambda=\lambda(u)$.
EDIT: Here's the easiest way to finish it up. Parametrize so that the $v$-curves are the rulings and the $u$-curves are the orthogonal trajectories (therefore also lines of curvature, so $n_u = kx_u$, where $k\ne 0$. Assuming the tangent planes all pass through the origin, we have $n\cdot x=0$, and so $n_u\cdot x = 0$, whence $x_u\cdot x =0$. It follows that $x=\lambda x_v=\lambda\beta$ for some scalar function $\lambda$. This tells us that our surface is a cone.
