Folding a Circle to Get an EllipseDate: 01/08/2001 at 09:35:01 From: Paul Rosin Subject: Folding circles to get ellipses I've read the following instructions: 1. Take a circle and place a point inside it. 2. Take a point on the circle and fold it to the interior point. Repeat this process, and the envelope of folds forms an ellipse. Can you tell me how to prove that the envelope is an ellipse? Also, what is its equation? Date: 01/08/2001 at 12:05:02 From: Doctor Peterson Subject: Re: Folding circles to get ellipses Hi, Paul. This is a nice construction, and is actually very closely related to the reflection property of an ellipse. I'll just give you a hint to get you started on a proof. Consider one of the lines you make by folding: ********* ****** ****** *** *** Q ** ******** ** A ****** /** * **+ / * * ** \ \ / * * ** \ \ / * * * \ /\ * * * \ / \ * * * \ / \ * * * +-------+-------------+ B * * O / C * * * / * * * / * * * / * * * / * * * / * ** * / ** ** * / ** *** * / *** ********/ ****** ********* P Here I've folded the circle on line PQ, so that point B on the circle lies on the fixed point A. You should immediately notice that PQ is the perpendicular bisector of AB; that's how folding works. Draw segment OB from the center of the circle, and look at point C where this intersects PQ. This will be the point on the ellipse. On one hand, you can show that point C fits the definition of an ellipse with O and A as foci; on the other hand, if you know that light from one focus reflects through the other focus, you should be able to see that PQ is in fact tangent to this ellipse. If you want the equation, you can use the foci and other information from this picture; or generate the equation from the construction by taking the angle of OB as a parameter, and constructing point C from point B. I haven't tried this, and can't tell you how hard it might be. If you need more help, feel free to write. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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