I have the following three points:
- A: One focus of an ellipse
- C: The furthest point from this focus (the far side of the major axis)
- D: Some other arbitrary point on the ellipse
From there, it's trivial to find:
- E: The "mirror point" of D across the major axis of the ellipse
I know it should be simple trigonometry from here to find B, the second focus of the ellipse, but I can't pull it off for whatever reason. Can anyone help me out here?
I'm working in polar coordinates and have the known focus of the ellipse at the origin.
Edit: Added point "E", which one can trivially find in my scenario. I think my scenario is distinct from this one, since we've added the restriction that one point is necessarily as far as possible from the focus along the major axis.
Edit 2:
I think I solved it. The distance between the two foci is given by: $$ \frac{4r_Cr_D-4r_C^2}{2r_D+2r_Dcos(\theta)-4r_C} $$ Where theta is the (positive) angle between C and D. I used law of cosines and the ellipse property if anyone wants to check my work.
In Cartesian coordinates, with the focus at the origin, the furthest point at $(c,0)$ and the other point at $(a,b)$, with $\sqrt{a^2+b^2}=r$.
The other focus is at $(x,0)$, and $$r+\sqrt{(a-x)^2+b^2}=c+c-x\\ (a-x)^2+b^2=(2c-x-r)^2$$ This is linear in $x$, so there will be one solution, provided $D$ is close enough.