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Given the general (real valued) equation of a conic section: $$ A x^2 + B xy + C y^2 + D x + E y + F = 0 $$ Then what is the circular cone associated with it ? Is it unique ? And is there a way to derive its equation, expressed in $(A,B,C,D,E,F)$ and $(x,y,z)$ ? I've done some homework here and here, but wasn't able to extract a simple method to find the cone(s), given the conic section.

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    $\begingroup$ Can you imagine many cones intersecting the $xy$-plane along the unit circle? All with the $z$-axis as the symmetry axis. The other conic sections have less symmetries, but I think we can still take advantage. After all, you can reflect the 3D-cone w.r.t. the plane giving this section. $\endgroup$ – Jyrki Lahtonen Aug 4 '14 at 10:47
  • $\begingroup$ You're quite right: a simple way to see that the solution is not unique. That's one thing settled. Thanks. $\endgroup$ – Han de Bruijn Aug 4 '14 at 10:49
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    $\begingroup$ I checked out your "homework". You may be interested in reading about Dandelin spheres. I'm not suggesting that they would help you solve this problem. Just that IMHO they are a pretty (and less well known) aspect of conic sections. $\endgroup$ – Jyrki Lahtonen Aug 4 '14 at 11:12
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    $\begingroup$ @JyrkiLahtonen: That's fun! The conic sections are introduced exactly in this way in an old schoolbook of mine, titled "beknopte Analytische meetkunde" written by D.J.E. Schrek (1964). The name of Dandelin is mentioned on page 83 of the book. (He published his discovery in 1822) $\endgroup$ – Han de Bruijn Aug 5 '14 at 19:37
  • $\begingroup$ Adding a link to this thread because... Well, nice pictures. $\endgroup$ – Jyrki Lahtonen Jan 14 at 9:38
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@Jyrki's suggestion to consider Dandelin Spheres is the key. It's possible (even easy) to construct a family of Dandelin Spheres from a particular conic, and these give the family of cones you seek.


Let's take the case of an ellipse. Viewing the curve's plane edge-on, we visually collapse the ellipse to its major axis $\overline{PQ}$. Let $F$ and $F^\prime$ be the foci.

Choose any $R$ such that $\overline{RF}\perp\overline{PQ}$, and draw the circle through $F$ with center $R$. This circle is the intersection of a Dandelin sphere with the plane perpendicular to the ellipse through its major axis.

enter image description here

Draw circles about $P$ and $Q$ through $F$ to determine points $S$ and $T$ on $\bigcirc R$. Necessarily, $\overleftrightarrow{PS}$ and $\overleftrightarrow{QT}$ are tangent to the circle. Let $C$ be the point where these lines meet.

enter image description here

The incircle of $\triangle PQC$ is our other "Dandelin circle".

enter image description here

From here, we see that $C$ is the apex the of the cone we seek. (If the tangent lines are parallel, then $C$ is the "point at infinity" and our cone is actually a cylinder.) Those tangent lines are the intersections of the cone with the perpendicular plane. Thus, we get a family of such cones based on the parameter-point $R$.

Finding the equation of the cone family should be relatively straightforward for an ellipse in "standard position". For a general ellipse, a few coordinate transformations will be helpful. We handle parabolas and hyperbolas similarly.

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My try.   Attempt to find an analytical solution. From the Conic Sections article as mentioned.
We have two sets of six variables: the well known $(A,B,C,D,E,F)$ for the conic section and the unknown $(\phi,\alpha,\gamma,p,q,h)$ for the cone. We also have six equations and three of them have been solved already in the article:$$ \tan{2\gamma} = \frac{B}{A - C} \\ \cos(\alpha) = \sqrt{\sqrt{B^2 + (A-C)^2}} \\ \cos(\phi) = \sqrt{\frac{(A+C) + \sqrt{B^2 + (A-C)^2}}{2}} $$ So we are left with three other equations - see article - and three unknowns $(p,q,h)$: $$ D = - 2 A\,p - B\,q + \sin(2\alpha)\cos(\gamma)\,h \\ E = - B\,p - 2 C\,q + \sin(2\alpha)\sin(\gamma)\,h \\ A p^2 + B p q + C q^2 + D p + E q + F = h^2 \left[ \cos^2(\phi) - \sin^2(\alpha) \right] $$ From the first two of these we find, with $h$ as the only unknown left: $$ p = \frac{- 2 C \sin(2\alpha)\cos(\gamma)\,h + B \sin(2\alpha)\sin(\gamma)\,h - B E + 2 C D}{-B^2+4 A C}\\ q = \frac{2 A \sin(2\alpha)\sin(\gamma)\,h - 2 A E - \sin(2\alpha)\cos(\gamma)\,h B + D B}{-B^2+4 A C} $$ Substitution of $p(h)$ and $q(h)$ into the third equation gives a quadratic equation in $h$, which can be solved, in principle. MAPLE does it in a page or two, but I find it not a pleasure nor instructive to reproduce them here. That's what I meant by not "a simple method"; hence the accepted answer.

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The cone is certainly not unique. For example, a circle is made by cutting a cone with a plane perpendicular to the cone's axis. But you could get the same circle by cutting a cone that has a smaller apex angle with a plane further from the apex (or whatever it is properly called).

You could ask for an equation with a parameter, giving a family of cones, but that is beyond me right now.

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  • $\begingroup$ Actually, I am asking for the latter: see the question. $\endgroup$ – Han de Bruijn Aug 4 '14 at 10:53
  • $\begingroup$ Yes, you are right, I did not see the "(s)" after "cone". Sorry! $\endgroup$ – Rory Daulton Aug 4 '14 at 10:54

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