Prove that a regular curve parametrized by arc length such that all its normal lines pass through a single point is contained in a circumference.
Suppose $\alpha:I\rightarrow \mathbb{R}^3$ is a curve such that all its normal lines pass through $P$. Then we can write that for every $s\in I$ there exists a $r_s$ such that:
$$\alpha(s)+r_s n(s)=P$$
or
$$r_s n(s)=P-\alpha(s)$$
then
$$\alpha'(s) \cdot (P-\alpha(s))=\alpha'(s)\cdot r_s n(s)=0$$
since $(P-\alpha(s))'=-\alpha'(s)$ which implies that the distance $|P-\alpha |$ is constant.
My question: how do I prove that $\alpha$ is in a circumference not in a sphere?
Thanks!
Best Answer
To show that the curve lies on a plane, it is enough to show that it is torsion-free (I'll let you figure out why) or that its binormal vector is constant (same). In your situation, the quickest way to show this is probably to use the Frenet-Serret formulas.