TITLE: What is the cone of the conic section?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.