That limit works just fine geometrically, keeping say the ellipse, and the angle between the cutting-plane and the cone's axis, constant. Then the ellipse is the intersection of the cylinder x^2 + y^2 = R^2 and the cutting-plane. It's then clear that the ellipse becomes the circular cross-section of the cylinder by squeezing uniformly in the direction of its major axis, but not at all in the direction of its minor axis. —Dan
On Jul 17, 2015, at 4:43 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Ooh. There is a continuum of cones that have a given ellipse as cross-section. 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?
Yes: https://en.wikipedia.org/wiki/Dandelin_spheres
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