1
$\begingroup$

Here's what I know:

  • The coordinates of a point A on an ellipse
  • The instantaneous slope of the ellipse at point A
  • Coordinates of one focus of the ellipse

Here's what I'm trying to find:

  • The equation of the ellipse (ideally parametric, but converting is easy)

The ellipse may or may not be at the center, and may or may not be rotated.

I want to know if this is possible - does the provided information define only one ellipse - and if the information defines only a few ellipses (not an infinite number) how could I get their equations from the known information.


What I've tried: I took the derivative of the equation of an ellipse, set it equal to the slope, plugged in the x and y of the point, and plugged in the focal length (substituting $a^2$ for $b^2+c^2$).

This works when the ellipse is not rotated and is at the origin, since the only unknown is $b$. After solving for $b$, you can plug it back in to the original equation and solve for $a$. When the ellipse is rotated and/or shifted from the origin, there are too many unknowns, though...

(This is my first post here, so let me know if I've made any mistakes!)

$\endgroup$
1
  • $\begingroup$ Have you tried anything to solve this problem? If so, it would be helpful to put it in your question so people can work off of that. $\endgroup$ Dec 24, 2016 at 23:18

3 Answers 3

3
$\begingroup$

As an illustration of the non-uniqueness of the stated conditions, the diagram below shows three distinct ellipses which share a tangent line, point of tangency, and focus. They are constructed in the manner described in JeanMarie's answer.

enter image description here

$\endgroup$
2
  • $\begingroup$ Did OP also imply the 3 major axes make same slope angle to tangent? $\endgroup$
    – Narasimham
    Dec 25, 2016 at 0:19
  • $\begingroup$ Not as far as I can tell from the OP. That's certainly not true in the picture I've drawn. $\endgroup$ Dec 25, 2016 at 0:20
1
$\begingroup$

These conditions do not define a unique ellipse [many thanks to @Semiclassical for pointing out an error in a prior answer].

In fact, due to the "reflection (or caustic) property" of the ellipse, the "ray" emitted from focus $F$ to the given point, say $M_0$, is reflected on the given tangent, and the 2nd focus $F'$ can be any point on this half line. Once you know the two foci of an ellipse and a point on it, the definition of an ellipse (set of points $M$ such that $MF+MF'=$ constant) gives a unique ellipse as solution.

$\endgroup$
5
  • $\begingroup$ You are perfectly right. I modify my answer. $\endgroup$
    – Jean Marie
    Dec 24, 2016 at 23:49
  • $\begingroup$ I believe this would only work with an ellipse that had not been shifted along the y axis, and had not been rotated? Imagine the ellipse had been shifted up so that the bottommost point was above the x axis. The intersection of the reflected ray and the x axis would be outside the ellipse! $\endgroup$ Dec 24, 2016 at 23:50
  • $\begingroup$ Oops, the other comments hadn't loaded... $\endgroup$ Dec 24, 2016 at 23:51
  • $\begingroup$ I've added an answer illustrating the construction you described. $\endgroup$ Dec 25, 2016 at 0:04
  • $\begingroup$ @Semiclassical Thanks. $\endgroup$
    – Jean Marie
    Dec 25, 2016 at 0:16
1
$\begingroup$

Also, I would say - these conditions do not define a unique ellipse

enter image description here

$\endgroup$
1
  • $\begingroup$ Nice, I think I like this picture better than mine. $\endgroup$ Dec 25, 2016 at 0:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.