Proving the existence of Dandelin Spheres

Question

I have been studying conic sections recently, and I came across the proof on how the algebraic definition of conic sections can be derived using Dandelin spheres.

I understood the proof for elipses, parabolas and hyperbolas. Where i am struggling is, that how can we be sure that a Dandelin sphere always exists. Is there a mathematical proof which ensures that a Dandelin sphere will always exist for all possible conics ?

I thought of proving this by proving that there always exists a unique sphere which passes through a circle and a point lying outside of the circle. But this doesn’t ensure that the sphere will be tangent along the circle to the cone and that it will be tangent to the plane (or does it ? If so, then I am unable to see it.)

Can you please help me out ?

Thanks in advance.

Given a cone and a plane $\alpha$, consider the plane $\beta$ passing through the axis of the cone and perpendicular to $\alpha$. That plane cuts the cone and $\alpha$ along a double angle and an intersecting line: the existence of Dandelin's spheres follows from the existence of the circles inscribed there, which is easy to prove. — Intelligenti pauca, Aug 05 '21 at 15:07
Tell me what you don't understand, then, and I'll try to explain it again. — Intelligenti pauca, Aug 05 '21 at 20:27
I understood that the intersection of those planes would be a line. But i don’t understand what you mean by ‘the plane cuts the cone and alpha along a double angle’. So i don’t understand how that leads to proving the existence of Dandelin sphere — Srinidhi kabra, Aug 06 '21 at 02:07
If you cut a double cone with a plane passing through its axis, the intersection is a double angle (i.e. a couple intersecting lines). — Intelligenti pauca, Aug 06 '21 at 06:40
Oh… i get it now… thank you very much for your explanation. — Srinidhi kabra, Aug 06 '21 at 07:19
Related: My answer to "What is the cone of the conic section?". — Blue, Aug 06 '21 at 07:19

score 1 · Accepted Answer · answered Aug 06 '21 at 07:10

Given a cone and an intersecting plane $\alpha$, consider the plane $\beta$ passing through the axis of the cone and perpendicular to $\alpha$. That plane cuts $\alpha$ at a line $AB$ and the cone at a couple of lines intersecting at the vertex $V$ of the cone ($VA$ and $VB$ in figure below, illustrating the case of an ellipse).

The axis of the cone is the bisector of $\angle AVB$. Construct then the bisectors of $\angle VAB$ and of its adjacent angle, intersecting the axis at $C$ and $C'$. As the bisector of an angle is the locus of points equidistant from the sides, points $C$ and $C'$ have the same distance from lines $VA$, $VB$, $AB$. Hence they are the centres of two circles tangent to those lines.

If you now rotate those circles about the axis of the cone, you'll get the desired Dandelin's spheres. I made the construction for the case of an ellipse, but the reasoning for a parabola or hyperbola is analogous.

Thank you very much. This clears things up for me. – Srinidhi kabra Aug 06 '21 at 07:19 — Srinidhi kabra, Aug 06 '21 at 07:19

Proving the existence of Dandelin Spheres

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