Analogue of a parabola on a sphere?

We can set this up with only a little algebra and trigonometry. Let $P$ be the north pole of the sphere $x^2+y^2+z^2=1$, also parameterized with polar angle $\theta\in[0,\pi]$ (where $\theta=0$ is the north pole), and radial angle $\phi\in[0,2\pi)$. The planes $x=0$ and $y\cos\alpha+z\sin\alpha=0$ are perpendicular, so the great circles $C$ and $D$ formed by their intersections with the sphere are also perpendicular. We want the angle from the north pole to equal the angle to $D$. The angle to $D$ is $\frac\pi2$ minus the angle to the perpendicular to $D$, and the cosine of this angle can be calculated by a dot product. The cosine of $\frac\pi2$ minus the angle to the north pole is easy to calculate; this is just $\sin\theta$. Thus we want:

$$\sin\theta=(0,\cos\alpha,\sin\alpha)\cdot(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)=\cos\alpha\sin\phi\sin\theta+\cos\theta$$

Solving this yields $\theta=\frac\pi2-\tan^{-1}(\csc\alpha-\sin\phi\cot\alpha),$ so a full parametric description of the "parabola" is this equation substituted into $(x,y,z)=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$.

The resulting figure is roughly circular, except near $\alpha=0$, when it narrows to an ellipse with foci at the pole and at the perpendicular to $D$.　Here's an animation for varying values of $\alpha\in(0,\pi)$:

EDIT: If we project this figure onto the plane $z=1$, we transform $(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)\mapsto(\cos\phi\tan\theta,\sin\phi\tan\theta,1)$, which is the equation for a figure in polar coordinates given by $r(\phi)=\tan\theta(\phi)$. Given our known value for $\theta$, this amounts to

$$r(\phi)=\frac{\sin\alpha}{1-\cos\alpha\sin\phi},$$

which is the equation of a (plane) ellipse. Thus this really is the intersection of an elliptical cone with the sphere.

We can set up the problem as an algebra problem.

Lines on the circle are great circles; we can choose our coordinates so that the line under question is the equator, and the point has coordinates $(0,a,b)$ with $a,b > 0$.

The closest point from $(x,y,z)$ to the equator lies in the same radial direction from the $z$ axis: that is, it is $\left( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}, 0 \right) $.

Thus, you seek the curve that simultaneously solves

$$ x^2 + (y-a)^2 + (z-b)^2 = \left(x - \frac{x}{\sqrt{x^2 + y^2}} \right)^2 + \left( y - \frac{y}{\sqrt{x^2 + y^2}} \right)^2 + z^2$$ $$ x^2 + y^2 + z^2 = 1$$