Geometry puzzle: plane X cylinder = ellipse

[belliott4488]

Actually, this is just a plain old geometry problem, really - no special tricks or anything. It just has a very nice solution, which someone showed me years ago (I take no credit).

Can you prove that the intersection of a right circular cylinder and a plane is an ellipse? (Assume the general case of an oblique intersection.)

There are probably plenty of ugly ways to do this; the point here is elegance!

 

[tiny-tim]

Is the answer you're looking for the one with the spheres?

 

[belliott4488]

Is the answer you're looking for the one with the spheres?

Yeah ... dang! I thought this was more obscure than it might be. Has everyone seen this before?

(Of course, it could still be considered "obscure" even if everyone on PF knew it!)

 

[tiny-tim]

It also works for cones, and with cones it works for parabolas and hyperbolas too.

 

[belliott4488]

It also works for cones, and with cones it works for parabolas and hyperbolas too.

Really?? ... Same kind of proof? I've got to try this! ...

 

[belliott4488]

Well, dog my cats! I never tried to do the proofs for the cases of intersections with a cone - that's very cool. The parabola case uses only one sphere, right? And you need the latus rectum - I believe it's called - but you can find that pretty easily.

Very nice - thanks.

 

[ManDay]

Okay, great you know how this works. Will you let the public (me!) know aswell?

 

[belliott4488]

Okay ... I'll be back a little later to type it up. Maybe I can draw a diagram, too, but that will take more time than I have right at the moment. Stay tuned.

 

[Werg22]

An interesting corollary, for lack of a better word, is that any parabola forms the part of an ellipse.

 

[belliott4488]

** SPOILER ALERT **
In case anyone has just come across this thread and still wants to take a shot at coming up with a proof, the following is the proof I had in mind, so skip it if you want to do it all on your own.

***************************

I'm not going to be rigorous, in the interests of saving space. In particular, I won't go through the details of constructions that I assume anyone reading this thread can probably come up with on his own.

Start with a right circular cylinder intersected at an oblique angle by a plane. We want to show that the intersection is an ellipse.

The first step is to construct two spheres, each with radius equal to the radius of the cylinder and center on the cylinder axis, so they will both be tangent to the cylinder. The centers of the spheres are located on the cylinder axis such that each sphere will be tangent to the intersecting plane, one above it and one below it. (This is the first construction the details of which I'll omit. It should be clear that such spheres can be defined and are unique.)

Now we consider a point P on the intersection curve of the plane and the cylinder. Construct a line from P to A, the point of tangency between the upper sphere and the plane. Next construct a line from P, parallel to the axis of the cylinder, and extend it to the circular intersection of the upper sphere and the cylinder. Call the point of intersection between this line and the circle U. Lines PA and PU are both tangent lines drawn from a point external to a sphere; therefore they are equal in length.

Next, construct similar lines PB and PL, from P to the point of tangency of the lower sphere and the plane, and from P to the circular intersection of the lower sphere and the cylinder, respectively. Again, we have PB=PL.

Now, since PU+PL is simply the distance between the circular intersections of the two spheres and the cylinder, this sum will be the same regardless of where the point P is chosen on the intersection of the plane and the cylinder. Since the PA=PU and PB=PL, however, we have also that PA+PB will be equal to PU+PL, and thus it, too, will be the same regardless of where the point P is chose.

Thus the intersection is an ellipse with foci at A and B.

Thanks to tiny-tim for pointing out that similar proofs can be constructed for the intersections of a plane and a right circular cone (all three cases, ellipse, parabola, and hyperbola). In these cases the circular intersections of the spheres and the cone will not be great circles, as they are for the cylinder, but this turns out not to matter. The case of the parabola is also somewhat different, since there will be only one sphere. In that case, the latus rectum of the parabola is constructed by finding the intersection of the first plane and the plane containing the circular intersection of the single sphere and the cone.