TITLE: Prove that normal lines to a cylinder cut the axis perpendicularly
QUESTION [1 upvotes]: Consider the following problem (Exercise 2.55 in Montiel and Ros's Curves and Surfaces, 2nd Edition):

Let $S = \{p \in \mathbb{R}^3 \ | \ |p|^2 - \langle p, a \rangle^2  = r^2\}$, with $|a|=1$ and $r>0$, be a right cylinder of radius $r$ whose axis is the line passing through the origin with direction $a$. Prove that $T_pS = \{v \in \mathbb{R}^3 \ | \ \langle p, v \rangle - \langle p, a \rangle \langle a, v \rangle = 0\}$. Conclude that all the normal lines of $S$ cut the axis perpendicularly.

Now, $S = f^{-1}(r^2)$ where $f(p) = |p|^2 - \langle p, a \rangle^2$. But 
$$
(df)_p[v] = 2 \langle p, v \rangle - 2 \langle p, a \rangle \langle a, v \rangle, 
$$
hence the first part.
Now, How to show the second part?
Any hints will be the most appreciated.

REPLY [1 votes]: Hint: Just observe that $a\in T_p(S)$ and the orthogonal projection  $p_{\perp a}:=p-\langle p, a\rangle\, a\, \perp T_p(S)$.
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