A curve in $\mathbb R^3$ starts at equator of sphere radius $a$ being inclined at $\alpha $ to longitude goes to North pole along a hyperbolic geodesic.
Find its radius $ r(\theta)$ as function of longitude $\theta$ in cylindrical/polar coordinates.
EDITS 1-3:
Considering differentials on a sphere meridian (arc of curve makes angle $\psi$ to longitude or sphere meridian) we have
$$ \sin \phi = \frac{ dr \tan \psi } { r d\theta}; $$
on the sphere of DE ($ \phi $ is slope angle between arc and symmetry axis)
$$ \cos \phi = r/a ;$$
but to proceed in finding parametric equations of the curve I am stuck not knowing dependence of $\psi $ on $r$... or $\theta$
Normally we have geodesics on a sphere as great circles, hyperbolic geodesics on a Beltrami pseudosphere as asymptotic lines of zero normal curvature.
But when seen interchangeably in the nature of geodesic, is it not possible making sense of it to parametrically define hyperbolic geodesics on a Riemann sphere and common geodesics on a pseudosphere?
At least we know geodesics cannot cross cuspidal equator of a Beltrami pseudosphere.
Now what about any restrictions that may exist in the present ( hyperbolic geodesics on sphere) case?
If not possible to define it, I appreciate your help to comment on how it fails to make enough sense. Regards