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In a project I'm working on the user creates a circle and choose a point on that circle, P=(px,py). For the question's sake, let's assume the center of the circle is at (0,0).

After the previous steps, the user can then change the eccentricity of the ellipse (as it was a circle it was actually an ellipse with e=0). While he changes the eccentricity, the ellipse should keep its center to (0,0), and the point P should stay on the ellipse's circumference.

Thanks! Aviad.

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This isn't a programming question! –  Will A Jul 17 '10 at 18:41
True. But I've seen a few mathematical questions laying around and figured it's okay to ask it... –  Aviad Ben Dov Jul 17 '10 at 18:42
Suggest mathoverflow.net –  Donnie Jul 17 '10 at 18:43
@Donnie: This question would be quickly closed on mathoverflow -- it's way too elementary for their audience of professional mathematicians. –  Jim Lewis Jul 17 '10 at 18:48
@Donnie : no, this is NOT for mathoverflow.net –  leonbloy Jul 17 '10 at 18:50

1 Answer 1

up vote 3 down vote accepted

If I made no mistake, the half axis of the ellipse are a = sqrt(x²+y²/(1-e²)) and b = a * sqrt(1-e²)

For the numeric eccentricity we have:

I) b = a * sqrt(1-e²)

and the equation for a point on the ellipse is:

II) x²/a² + y²/b² = 1

Substitue I) in II)

x²/a² + y²/(a² * (1-e²)) = 1

1/a² (x² + y²/(1-e²)) = 1

a² = (x² + y²/(1-e²))


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+1, I get the same result. –  Jim Lewis Jul 17 '10 at 19:09
x,y in this answer refer to my px, py, right ? i.e. the point in the original circle? –  Aviad Ben Dov Jul 17 '10 at 19:14
x and y refer to all points on the ellipse, and of course px, py is one of them :-) –  Landei Jul 17 '10 at 20:29
Yeah, I figured that. :) it worked perfectly. Thanks! –  Aviad Ben Dov Jul 17 '10 at 20:39

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