calculate ellipse tangential to 2 other ellipses sharing 1 focus

Question

Problem

As illustrated here, i have two ellipses ($E_0$ aand $E_1$) both with $(0,0)$ as their one focus point.
The other focuspoint ($f_0$) of $E_0$ lies on the negative side of the x axis per definition, but the other focuspoint ($f_1$) of $E_1$ can lie anywhere.
I do also know the major and minor axie of both ellipses, in this particular illustration $E_0$ lies inside $E_1$ but this is not necesarrily always the case.
However it is per definition always true that $E_0$ and $E_1$ never intersect oneanother.

Given a specific point $A$ on $E_0$ i want to draw a third ellipse $E_2$, whose one focus also must be at $(0,0)$ and which is tangential to $E_0$ exactly at $A$ and tangential to $E_1$ at any point.

My solution this far

This problem really comes down to finding the other focus ($f_2$) of $E_2$.

By doing some experiments i have found that in order to be tangential to $E_0$, $f_2$ must lie on a ray from $A$ through $f_0$, (which also implies that $B$ must lie on the intersection between $E_1$ and a ray from $f_1$ through $f_0$) – this can also be seen in the illustration.
I have not been able to come up with a proof for this, but since it seems to be the case in all the experiments i have made, i assume it to always be true.
Therefor the problem simplifies to finding the correct distance $l$ between $A$ and $f_2$.

I have been able to find a way of calculating the $B$ (as the intersection between $E_1$ and the ray from $f_1$ through $f_2$) from a known $l$ (which is descriped later).
When i do have $B$ i can calculate the distances from $B$ to $(0,0)$ and from $B$ to $f_2$ (which easilly can be calculated from l), only if the sum of these to distances is equal to the sum of the distances from $A$ to $(0,0)$ and from $A$ to $f_2$ does $B$ lie on $E_2$, which (if my previous assumption is true) implies that the $E_2$ is tangential to both $E_0$ and $E_1$.

It also appears that as $l$ aproached the correct value, the difference between the previously mentioned sums aproach $0$ (no matter if $l$ aproaches the correct value from a greater or smaller value; therefor i have this far been able to recursively (since this is all a part of a simulation which i am programming) find an almost acceptable aproximation for $l$.

There are two problems with this: firstly it is only an aproximation, and secondly it is recursive, and even though it is acceptably accurate, and acceptably fast, i do still not like the idea of using a recursive aproximation if a way of calculating $E_2$ correctly exist.

Calculating $B$

This is way i have been able to calculate the point $B$, though i can not say for sure that a more logical and simple aproach to this could exist.

Firstly, to easilly find the intersection between $E_1$ the ray from $f_1$ through $f_2$, i transform $E_1$, $A$, $f_0$ and $f_1$ so that $E_1$ can be defined as $1=\frac{y^2}{b^2}+\frac{x^2}{a^2}$, this is done by multiplying all points and direction with a four dimensional matrix which i have in my simulation, and which beyond all reasonable doubt transforms correctly (my specific matrix always transforms $f_1$ to the negative side of the x axis and the focus which before was at (0,0) to the positive side of the x axis).

The point $f_2$ then becomes $\vec{f_2}=|\vec{a}-\vec{f_0}|l$ (In my simulation the points are saved as vectors, therefor this is valid) the direction of the ray from $f_1$ through $f_2$ is then $\vec{ray}=\vec{f_1}-\vec{f_2}$.
Then the $y=s\cdot x+m$ eqaution for this ray can be found, since $s=\frac{\vec{r}_y}{\vec{r}_x}$ and $m=\vec{f_1}_y-s\cdot\vec{f_1}_x$.

Then finding the x coordinate of $B$ is simply a matter of finding the intersection between an ellipse defined by $1=\frac{y^2}{b^2}+\frac{x^2}{a^2}$ and a ray with a known equation, which comes down to solving this equation of the second degree: $0=x^2\cdot (\frac{s^2}{b^2}+\frac{1}{a^2})+x\cdot( \frac{2\cdot s\cdot m}{b^2})+\frac{m^2}{b^2}-1$.
This does yield two solutions, but if $s$ is positive, then the smallest sollution is true, otherwise it is the greatest which is true (in my case, where the transformation matrix transforms $f_1$ to a negative x value on the x axis).
The y coordinate can then be found by inserting the found x coordinate into the eqaution of the ray $y=s\cdot x+m$.

Finally the coordinates of $B$ can be transformed back to the original space, by using the inverse transformation matrix.

question specification

My question is if there is a way of calculating the correct position of $f_2$ (without recursively aproximating it as i do) so that $E_2$ can become tangential, under the conditions outlined in the »problem« section. If this is possiple i would like to know how.

No matter if it is possiple or not, i would – if possible – like to see a proof of why it is so.

Answer 1

Shortly after my own question, someone else posted a similar problem, but where the first two ellipses do intersect oneanother, in this question the answer posted by Jack D'Aurizio was that the correct other focuspoint of the transfer ellipse, always lie on a hyperbola through a point I on the line between $f_1$ and $f_0$, where it is true that $|f_1I|+|f_0I|=a_{E_1}-a_{E_0}$ (where $a$ is the major axis of the ellipses, and $|f_1I|$ and $|f_0I|$ is the distances between I and $f_1$ and $f_0$)

This is however only true when the ellipses intersect oneanother, as ccorn correctly pointed pointed out in a comment, in my case where the ellipses don't intersect (or as he said $|a_{E_1}-a_{E_0}|>|f_0f_1|$ then the correct $f_2$ lies on an ellipse, with focuspoints $f_0$ and $f_1$ and through $I$

The specific point $f_2$ is furthermore exactly the one of the intersections between the ray from $A$ through $f_0$ and the previously mentioned ellipse, which lies furtherst from $A$.