What is the cone of the conic section? 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.

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.
