Parabola: the set of points in the plane that are equidistant from a line, called the directrix, and a point, called the focal point, not on the line.

Suppose we try to replicate this on a sphere: let the directrix be given as a great circle $D$. Let $C$ be a great circle perpendicular to $D$. Let $P$ be our focal point, lying on $C$ (but not lying on the directrix $D$). Let e be an arbitrary point on $D$, and let $E$ be the great circle passing through e, perpendicular to $D$. Let a great circle through $P$ meet $E$ at point f such that f is equidistance from e and $P$. Finally, let our "parabola" consists of all such points equidistant from focal point $P$ and directrix $D$ (measured along the "straight lines" of great circles).

From what I can tell, there are only two possibilities for the shape of this "parabola": either (1) $P$ will lie on one of the apexes of the circle (that is, one of the two points where great circles perpendicular to $D$ intersect), in which case the "parabola" will be a small circle; otherwise, (2) $P$ will lie at some other point along $C$, in which case I'm uncertain about the shape, other than that it will be symmetrical with respect to $C$ and occupy one hemisphere (since it will lie entirely above $D$).

What will be the shape of the curve under condition (2)?

  • $\begingroup$ This strikes me as a pretty interesting question which might be handled well by some curve/surface differential geometry in $\Bbb R^3$! $\endgroup$ – Robert Lewis Apr 6 '14 at 3:51
  • $\begingroup$ @Ryan Cool, great question! $\endgroup$ – user98602 Apr 6 '14 at 3:54

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:


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


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

  • $\begingroup$ Looks very nice, +1! $\endgroup$ – Robert Lewis Apr 6 '14 at 7:20
  • $\begingroup$ @1950RobertLewis Updated due to incorrect derivation. (I calculated the point halfway between the focus and the line for various points on the line; doing this in euclidean space would just get a line halfway between the directrix and the focus.) $\endgroup$ – Mario Carneiro Apr 6 '14 at 9:31
  • 1
    $\begingroup$ +100 if I could. I'm unfamiliar with spherical coordinates, so having the animation there to reify that math really makes this answer come alive for me. Thanks so much. $\endgroup$ – Ryan Apr 6 '14 at 14:56
  • $\begingroup$ Do you have a simple description of the shape? Is it actually an ellipse in 3-space? I suspect it's the intersection of a cone (maybe an elliptical cone) with the sphere, but I haven't ground out the details yet. $\endgroup$ – user14972 Apr 6 '14 at 15:06
  • 2
    $\begingroup$ @Hurkyl I just checked, and it really is the intersection of an elliptical cone with the sphere. See my edit. (It is not an ellipse in 3-space, because it is clearly non-planar - the only planar figures on a sphere are circles.) $\endgroup$ – Mario Carneiro Apr 7 '14 at 5:49

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.