Proof that an Ellipse exists


          Although we have the string, Graphing calculator, and GSP to show us that we’ve created an ellipse how can we be sure that our drawings are indeed a true ellipse? Well, I’m not sure how we could provide an elegant method for the ‘String’ proof or a concise method for our Graphing calculator, but I am positive that with GSP I can give a very nice proof. Here are a couple of things to consider that I may not have gone over. First of all the distance from the edge of the ellipse to the first focus plus the distance from the same point on the edge of the ellipse to the second focus is always a constant distance. Moreover wherever you place this point on the outer edge of the ellipse will have this same property. If you’re not convinced then click here for a GSP visual confirmation. Here’s a visual for those of you that don’t have GSP.

          The second property that may be of importance in this proof is the fact that if an object such as a sound wave or a ray of light passes through one foci then it will reflect off the inside of the ellipse and at least pass through the opposite foci. Here’s a picture to demonstrate.

          Suppose our picture signifies a real ellipse with an opening starting at the dashed line and having a reflective property to the left of the dashed line. Then as the starting point of light ray goes through F1 and hits the reflection point it will be shot directly to F2 and continue going away from F2 for ever into infinity.

          So with the help of those two important properties we can discover a proof for the presence of an ellipse. Although we may trust GSP and our eyes we still may see something that actually isn’t there at all. So let’s begin with the proof. First we need to label our points to make an elegant and concise proof.

If we say that A and B are the foci of the ellipse then by the definition of an ellipse the distance from A to some point C on the ellipse plus the distance from B to the same point C is a constant. We begin the proof by saying that BD is a constant because it is the sum of BF (our large circle’s radius) plus FD (our small circle’s radius, which is equal to GA). So the first part of our ellipse, BC, is just BD minus CD, which turns out to be the radius of our smaller circle plus the radius of our tangential circle. This length, CD, is not constant, but fortunately for us it is congruent to some other piece of our picture. It turns out that triangle ACE is congruent to triangle DCE, and I’ll show you why. We constructed segment AD, found the midpoint, and called it E. So E bisects AD and therefore AE is congruent to DE. We also constructed the line through EC by dropping a perpendicular with AD through E, which makes both angles AEC and DEC right angles, and hence these angles are congruent also. Finally both triangles share a side CE. So by the side-angle-side axiom we can say that triangles ACE and DCE are congruent. Therefore AC must be equal to CD. And as we said before BC plus CD is a constant and by our definition of an ellipse, and thanks to the SAS axiom we can say that AC + BC is a constant length.  Through all this we learn that our locus of construction is in fact an ellipse as we had theorized.