Let's say I have an ellipsoid centered at the origin of my coordinate system, and not rotated in any way. Think along the lines of the WGS84 ellipsoid in ECEF. Let's call this ellipsoid E1.
Now, let's say I pick a point P1 on the surface of E1, and I pick two points P2 and P3 outside E1. I now construct a second ellipsoid E2 with P2 and P3 as the foci, such that the surface of E2 contains P1.
I can see, qualitatively, that E2 will intersect E1. E2 is not centered at the origin, and may be 'rotated'.
My question is: how do I analytically compute that intersection? Either via a single expression or a series of transformations?
My first thought was to hypothesize a plane containing the intersection, and computing that plane as a minimization problem with E1 and E2 as constraints. I am not sure if this is correct or even how to set up the problem.
Any help with this would be greatly appreciated. Thanks!
By a suitable affine transformation, you can turn one of the ellipsoids to a unit sphere, while the other is centered on the origin and axis-aligned.
The implicit equations are $$(x-u)^2+(y-v)^2+(z-w)^2=1,$$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$$
You can use the parametric equations of the sphere and plug into the ellipsoid to get a compatibility condition between the angles:
$$\frac{(\cos\theta\sin\phi+u)^2}{a^2}+\frac{(\sin\theta\sin\phi+v)^2}{b^2}+\frac{(\cos\phi+w)^2}{c^2}=1.$$
Unfortunately, there will be no nice simplification and you end-up with a nasty quartic trigonometric polynomial.
Note that you can expand the equation in terms of one angle and obtain a form like
$$p(\phi)\cos^2\theta+q(\phi)\cos\theta\sin\theta+r(\phi)\sin^2\theta+s(\phi)\cos\theta+t(\phi)\sin\theta+u(\phi)=0.$$
Then by the transformation
$$\cos\theta=\frac{t^2-1}{t^2+1},\sin\theta=\frac{2t}{t^2+1},$$
you obtain a quartic polynomial, for which a (heavy) closed-form solution is possible.