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Rebesques · #2 June 7th, 2009, 11:26 PM

a) Let the curve be parametrized by arclength, $x=x(s)$ and let $\{t,\eta,b\}$ be the Frenet-Serret frame. Then the plane $\{\eta,b\}$ always passes through a point $P$ . Express $P=\lambda(s)\eta(s)+\mu(s)b(s)$ and differentiate to obtain $\lambda, \mu=0$ . So $\{\eta,b\}$ always crosses the origin, which means $x\in\{\eta,b\}$ or $x(s)=\Lambda(s)\eta(s)+M(s)b(s)$ . Use the Frenet-Serret equations to show that $\Lambda, M$ are constants.

(Actually, the curvature of $x$ turns out to be constant and the torsion zero, so $x$ is a circle.)

b) Let the coordinates be x,y,z. We easily see that this circle is the intersection of the circular cylinder $x^2+z^2=1$ with the parabolic cylinder $y^2=2(1+x)$ .
(...unless i messed up my calcs again)