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I have a rotated ellipse, not centered at the origin, defined by $x,y,a,b$ and angle. Then I have a segment defined by two points $x_1$, $y_1$ and $x_2$, $y_2$. Is there a quick way to find the intersection points?

M47145
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ilbiffi
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  • What do you mean by easy? Substitution should be easy. – hjhjhj57 Sep 04 '14 at 06:26
  • I mean by as few steps as possible. – ilbiffi Sep 04 '14 at 10:27
  • These equations

    http://mathworld.wolfram.com/Ellipse-LineIntersection.html

    work well but only for an ellipse that is placed at 0,0 and also if a line is passing through the center. What I was looking for was a more general equations for the intersection, for a generic ellipse centered everywhere and arbitrarily rotated and for a generic line not passing through 0,0.

     – ilbiffi Sep 04 '14 at 11:38
  • I used wolfram alpha equation solver, I tried to insert the equation of a line into the one of a standard non rotated, non translated ellipse,

    http://i58.tinypic.com/10how40.gif

    and resolving for x this is the result

    http://i62.tinypic.com/15qw704.png

    which is nice.

    Then I took the equation of a rotated and translated ellipse

    http://i57.tinypic.com/vngnwi.png

    and this is the result

    http://i58.tinypic.com/2rnzx1h.png http://i57.tinypic.com/6p5bat.png

    Which is obviously impractical, can anyone suggest a different method?

     – ilbiffi Sep 05 '14 at 19:40
  • I think the best you will be able to do is get to the unrotated case. For which you would only need to find the eigenvectors of the ellipse. I think they work an example here: http://en.wikipedia.org/wiki/Principal_axis_theorem. You should also check http://math.stackexchange.com/questions/761460/rotation-of-conics-sections-using-linear-algebra – hjhjhj57 Sep 05 '14 at 20:25

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