On surfaces of revolution (II)

The idea about the concept of a surface of revolution is as 'something that rotates'. However there is a characterization of this class of surfaces in terms of the tangent planes. It is not difficult to see that in a surface of revolution the normal lines through any point meets the rotation axis. Now, and it is here the surprise, this property characterizes a surface of revolution. Thus the result is the following.

Theorem. If all normal lines in a surface meet a given straight-line $L$, then the surface is included in a surface of revolution and $L$ is the rotation axis.

The proof consists into prove that the intersection of any orthogonal plane to $L$ with the surface $S$ is a circle centered at $L$. Then it suffices to finish the result: the surface is formed by the union of (arcs of)  circles centered at $L$ and this is just the definition of a surface of revolution. Denote by $N$ the unit normal vector to $S$.

First step. Let $P$ be a orthogonal plane to $L$ that meets $S$. In particular, for any $p\in S\cap P$, $P\not= T_pS$: on the contrary, the normal line is parallel to $L$ so it does not meet $L$. Thus  $S$ and $P$ meet transversally and $S\cap P$ can parametrized as a regular curve, namely, $\alpha=\alpha(s)$. 

Second step. The normal line of $\alpha$, as curve of ${\mathbb R}^3$, meets $L$. First, recall that the normal line is included in $P$. Furthermore, the normal vector $n(s)$ of $\alpha$ at $s$ is orthogonal to $\alpha'(s)$, which lies in $P$. But the orthogonal projection $\pi(N(\alpha(s)))$ of $N(\alpha(s))$ on $P$ is also a vector orthogonal to $\alpha'(s)$. Thus $n(s)$ and $N(\alpha(s))$ are collinear. Since the normal line through $\alpha(s)$ meets $L$, the same occurs for the line through $\alpha(s)$ and with direction $\pi(N(\alpha(s)))$. 

Third step. The only planar curve whose normal lines meet at one point $p_0$ is a circle centered at $p_0$. Indeed, for each $s\in I$, there exists $\lambda(s)$ such that $p_0=\alpha(s)+\lambda n(s)$, where we are assuming that $\alpha$ is parametrized by the length-arc. If we differentiate with respect to $s$ and using the Frenet equations, we obtain, $$0=\alpha'(s)+\lambda'(s)n(s)-\lambda(s)\kappa(s)\alpha'(s).$$ This proves that $\lambda'=0$ on $I$, that is, $\lambda$ is a non-zero constant and $1-\lambda(s)\kappa(s)=0$, so $\kappa(s)=1/\lambda$, that is, $\alpha$ is included in a circle.