Ellipse like on sphere

Find the locus of all points on a sphere such that the sum of geodesic distances from two fixed points $F_1$ and $F_2$ on it is a constant, less than its diameter. ( When radius of sphere goes to infinity, it would look like an ellipse).

Following is image after Equn ($1$) in achille hui note

Solution 1:

WOLOG, consider the case $\begin{cases} F_1 &= (+\sin\alpha,0,\cos\alpha),\\ F_2 &= (-\sin\alpha,0,\cos\alpha) \end{cases}$ with $\alpha \in (0,\frac{\pi}{2})$ and the sphere is the unit sphere.

Let $2\theta \ge 2\alpha$ be that constant for the sum of geodesic distances to $F_1$ and $F_2$. For any point $p = (x,y,z)$ on the locus. we can find a $\tau$ such that

$$ \theta + \tau = \text{dist}(p,F_1) = \cos^{-1}(\cos\alpha z + \sin\alpha x)\\ \theta - \tau = \text{dist}(p,F_2) = \cos^{-1}(\cos\alpha z - \sin\alpha x) $$ This leads to $$\cos(\theta\pm\tau) = \cos\alpha z \pm \sin\alpha x \implies \begin{cases} \cos\theta\cos\tau &= +\cos\alpha z\\ \sin\theta\sin\tau &= -\sin\alpha x \end{cases} $$ and hence $$\left(\frac{\cos\alpha}{\cos\theta} z\right)^2 + \left(\frac{\sin\alpha}{\sin\theta} x\right)^2 = (\cos\tau)^2 + (\sin\tau)^2 = 1$$

Let $A = \tan\alpha$ and $D = |\tan\theta|$, we can rewrite this as

$$z^2 + \frac{A^2}{D^2} x^2 = \frac{1+A^2}{1+D^2} \quad\iff\quad \frac{x^2}{D^2} + \frac{y^2}{D^2-A^2} = \frac{1}{1+D^2}\tag{*1} $$

$\require{enclose}\newcommand{\mybox}[1]{\enclose{roundedbox}{\;#1\;}}$ Geometrically, there are several possibilities

• $\mybox{\theta = \alpha}$

  $D = A$, $(*1)$ reduces to $y = 0$ and $|x| \le \cos\alpha$. The locus is a circular arc joining $F_1$, $F2$.

• $\mybox{\alpha < \theta < \frac{\pi}{2}}$

  $A < D < \infty$, $(*1)$ reduces to the equation of an ellipse. The locus lies on the upper-hemisphere. It not only looks like an ellipse, it is an ellipse if you project it to the $xy$-plane.

• $\mybox{\theta = \frac{\pi}{2}}$

  $D = \infty$, $(*1)$ reduces to $x^2 + y^2 = 1$. The locus is the equator.

• $\mybox{\frac{\pi}{2} < \theta < \pi - \alpha}$

  $A < D < \infty$ again, the locus lies on the lower-hemisphere. Once again, it is an ellipse if you project it to the $xy$-plane.

• $\mybox{\theta = \pi -\alpha}$

  $D = A$ and once again, $y = 0$ and $|x| \le \cos\alpha$. The locus is a circular arc joining the antipodal points of $F_1$ and $F_2$.

Update

At the end is a picture showing how the family of loci qualitatively look like. The plot is generated for $\alpha = 30^\circ$ with $\theta$ start at $30^\circ$ increasing with step $5^\circ$. The loci whose $\theta$ is an integer multiple of $15^\circ$ is colored in black and the one for $\theta = 90^\circ$ is in white.

Please note that the plot is partially transparent. If you look carefully, you can see the locus for $\theta = 90^\circ$ behind the back of sphere. This locus doesn't have any strange feature and it is simply the great circle at equator!

$\hspace1.0in$

Solution 2:

In response to queries by ebin zheng:

Achille hui has taken unit radius for the sphere. However if we denote it by $a$ we have

$$ \frac{x^2}{D^2} + \frac{y^2}{D^2-A^2} = \frac{a^2}{1+D^2}\tag{**1}$$

and you wish to recover planar ellipse as $ a \rightarrow \infty. $