On 2015-07-18 05:11, Fred Lunnon wrote:
Thermally challenged theorem!
[However, note what happens where the second conic meets the plane of the first.]
WFL
The two curves remain in perpendicular planes. The family of cones through each curve is one-dimensional, but X-shaped. I.e., when a cone goes flat, it can instantaneously switch its apex to lie on the other branch of its dual curve. The two cones independently oscillating around planarity and vacillating between branches might make an interesting animation. --rwg
On 7/18/15, Bill Gosper <billgosper@gmail.com> wrote:
On 2015-07-17 16:43, Allan Wechsler wrote:
Ooh. There is a continuum of cones that have a given ellipse as cross-section.
This seems a fine time to reprofess that the apices of these cones all lie on the same hyperbola, and the apices of cones through that hyperbola all lie on your original ellipse. gosper.org/conethm.png Each curve contains the other's foci. --rwg
In the limit, with the apex of the cone at infinity, the cone becomes a cylinder. Intuitively, the Dandelin construction must still work with a cylinder. And proving that the cross-section of a cylinder is a stretched circle seems like it should be easy. I bet there is a good proof hiding here.
On Fri, Jul 17, 2015 at 5:04 PM, Andy Latto <andy.latto@pobox.com> wrote:
That's a great geometrical proof that "Set of points with a constant distant sum from two foci" is the same as "intersection of a circular cone and a plane". But what's the geometrical argument that either of those is the same as a stretched circle?
Andy
On Fri, Jul 17, 2015 at 3:05 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Is there a proof this short and simple that the ellipse defined by these definitions is related to a circle by an affine transformation?
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