Ellipse tangential to two arbitrary ellipses sharing a focus

Question

In orbital mechanics, a bi-tangential transfer uses an elliptical orbit touching two other orbits. They are generally very efficient.

Suppose you have two arbitrary planar ellipses, sharing a focus $O$. Their second foci are $A$ and $B$. A bi-tangential ellipse would be a third one, also sharing the common focus point, and is is tangential to both the ellipses. It has its own second focus, $P$.

For any given pair of orbits, there are an infinite amount of such bi-tangential ellipses. The tangential points are where a ray from $P$ through $A$ or $B$ crosses the perimeter of the ellipses.

All possible locations of point $P$ are inside both ellipses, along a straight line. This line is a normal to the line $AB$, and is crossing it in $I$. Below is an illustration of the situation:

How can I, given two ellipses, calculate the location of $I$?

This unanswered question, is very related, showing way more detail up til the point of finding the rays with $P$, $A$ and $B$, and the tangential points. It however never finds the line of possible $P$ locations, instead falling back to a numerical algorithm.

Answer 1

Lovely question!

Assume that the black ellipse $\mathcal{E}$ meet $\mathcal{E}_B$ at $P_B$ and $\mathcal{E}_A$ at $P_A$ only. Then the other focus of $\mathcal{E}$, $P$, has to lie on the $BP_B$ line and on the $AP_A$ line. We have $PP_A+OP_A = PP_B+OP_B$ and $AP_A+OP_A$ equals the length $a_A$ of the major axis of $\mathcal{E}_A$. It follows that $PA-PB$ equals $a_B-a_A$. This holds also in the particular case in which $P_B=AB\cap\mathcal{E}_B$. In such a case $P_A=AB\cap \mathcal{E}_A$, so

$I$ can be found by locating the only point of the $AB$ segment such that $IA-IB=a_B-a_A$.

However, it is not true that the locus of points $P$ is a line through $I$: instead, it is a hyperbola through $I$ (and through the intersections of $\mathcal{E}_A$ and $\mathcal{E}_B$), having its foci at $A$ and $B$: