Been attempting again to find a neat equation on a unit radius sphere for Sphere Ellipse locus in 3D conceptually similar to a plane ellipse.
Geodesic arc distances between point P, in spherical coordinates $(r=1,\theta,\phi)$ and two foci on either side of equator $F_1,F_2= (1,0,\pm c)$
Distances $(PF_1,PF_2)=(d_1,d_2) $ using Haversine formula for geodesic arc distances on sphere have constant focal lengths sum
$$d_1+d_2= 2a >2c \tag1$$
( we can allude to "eccentricity" $ = e^*= c/a $ even in 3D )
$$ (1-\cos d_1)= ( 1-\cos( \phi -c))+\cos^2\phi\, (1-\cos \theta)\tag2$$ $$ (1-\cos d_2)= ( 1-\cos( \phi +c))+\cos^2\phi \,(1-\cos \theta)\tag3$$
Parameterization tried as. $d_1=a+h, d_2=a-h\, $ Subtracting 2) from 3)
$$ \cos a \cdot \cos\ h = \sin \phi \sin c \tag4 $$
$$ \cos (a+h) = \cos( \phi -c)- \cos^2\phi\, (1-\cos \theta)\tag5$$
$h$ can be eliminated from above two to get $ F (\phi, \theta) = 0$ Is there some equation more elegant?
How to proceed finding other elegant implicit $F(\theta,\phi)=0 \,$ or explicit $\,\phi=f(\theta)$ or a parametric equation $ \theta(t), \phi(t) $ for this ellipse-like locus?
Another relation which may or may not be of use in finding the equation with dihedral $\gamma$ opposite to segment $F_1 F_2$ on prime meridian is
$$ \sqrt{ d_1d_2}\,\cos{\gamma/2} = \sqrt{a^2-c^2} \tag6 $$
Thanks for help to take it further.
From the above the loci are drawn on unit sphere using parameter $h$.
There are two "sphere ellipse" sets and one "sphere hyperbola" set (that was not expected at start), separated by two intersecting great circles (not drawn ). Drawn to $c=0.3,a$ values in interval $(0.4,2.8,0.2)$.