Proove locus of foot of perpendicular lies on hyperboloid

Question

Find the curvature vector and it's magnitude at any point r(t) of the curve r =( a cost, a sint, at). Show that locus of foot of perpendicular from origin to the tangent is a curve that completely lies on the hyperboloid $$x^2 + y^2 -z^2 = a^2$$. How to do the second part of the question??

Answer 1

HINT

consider a point $$\vec r_0=(a \cos t_0, a \sin t_0,at_0)$$

and the tangent at this point $$\vec t_0=(-a \sin t_0, a \cos t_0,a)$$

consider the line

$$\vec r(s)=\vec r_0 +s\cdot\vec t_0$$