The equation of a "standard" circle is $x^2 + y^2 = r^2$. That equation is not homogeneous and does not include the origin; we can homogenize it by adding a $z$ term of degree 2, getting $x^2 + y^2 = z^2$, a cone.
Given $x^2 + y^2 = z^2$, we can retrieve the circle of any $r$ by intersecting with $z = r$.
What if we want to do something similar for conics (not just circles)? Can we develop a homogeneous equation which allows us to take not only circles but any conics?
I imagine we would need to do something like $ex^2 + ey^2 = ez^3$, where $e$ is eccentricity. However, I do not know how to complete this approach.
Another approach is to work with the original equation, but instead of solving for $z = r$, solve for $z = r + kx + jy$, where $k$ and $j$ are arbitrary constants. This seems more promising. How would I express $k$ and $j$ in terms of eccentricity? And, since I don't care about rotation, can I just set $j = 0$, yielding $z = r + kx$? How does this relate to eccentricity?
Note: There are many references giving the general equations for conics. However, these include translation, rotation, and sometimes skew. I'm looking for a simple homogeneous equation which produces only "standard" conics, that is, intersecting a plane through a cone. And, I'd like to be able to find the eccentricity, ideally having it appear in the equation.
Update
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A homogeneous equation for one non-homogeneous parameter, the eccentricity $e,$ in two cases $e>0$ or $e=0:$ $$(F_z x-F_x z)^2+(F_z y-F_y z)^2=(A x+ B y+ C z)^2, e>0 $$ or $$(F_z x-F_x z)^2+(F_z y-F_y z)^2= (r z)^2, e=0, r>0.$$ To get back the metric properties, set $(F_x:F_y:F_z)=(f_x:f_y:1), z=1$ where $(f_x,f_y)$ is the the focus for the corresponding directrix $a x+b y+ c=0,$ with $a,b,c$ s.t $(A, B, C)=(ea, eb, ec), a^2+b^2=1.$ Note that we almost could unify the two cases with the circle being for $A=B=0,C=r,$ making $z=0$ (the line at infinity) the directrix.
As examples: a parabola $y-x^2=0:$ $x^2+(y-z/4)^2=(y+z/4)^2$
an ellipse $x^2+y^2/4-1=0:$ $x^2+(y+\sqrt3 z)^2=\frac34 (y+4z /\sqrt3)^2$ or $x^2+(y-\sqrt3 z)^2=\frac34 (y-4z /\sqrt3)^2$
a hyperbola $x^2-y^2/4-1=0:$ $(x+\sqrt5 z)^2+y^2=5 (x+z/\sqrt5)^2$ or $(x-\sqrt5 z)^2+y^2=5 (x-z/\sqrt5)^2$
Note that when you have two foci there is one representation of the equation for each focus.