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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.

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geofurb
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  • Yeah, this might be the same problem... I just ran the trig again and I'm getting that the distance between the two foci is not constrained. – geofurb Apr 13 '16 at 05:05
  • Added the length constraint from the definition of an ellipse and that's given me something to work with. Looks good? – geofurb Apr 13 '16 at 05:54

1 Answers1

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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.

Empy2
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