Equations for infinite double cones in the Cartesian, cylindrical and spherical coordinate systems.

Cartesian

We start with the equation for a circle centered at the origin, but instead of treating
the radius as a constant, we make it a variable in the third dimension.
$$x^2 + y^2 = z^2$$
All points \((x, y, z)\) that satisfy this equation are on a double cone.
We can picture the double cone as an infinite number of circles parallel to the x-y plane.
For each radius \(\sqrt{x^2 + y^2}\) there are two circles, one at \(z = \sqrt{x^2 + y^2}\) and one at \(z =
-\sqrt{x^2 + y^2}\).
If we make a cut at \(y = 0\), we get the following shape.

Another way of conceptualizing the double cone is as a line \(z = mx\) with slope \(m = \tan \alpha\) rotated about the z-axis.
Our above equation uses the radius for z without any scaling, so \(m = 1\).
But in general, we have
$$
(mx)^2 + (my)^2 = z^2.
$$

Cylindrical

To get the cylindrical coordinates from Cartesian coordinates, we simply replace the x- and y-coordinates with polar coordinates and keep z.
Each point then is described with a triple \((r, \theta, z)\), where \(r^2 = x^2 + y^2\) and \(\tan \theta = y/x\).
That's very convenient for our double cone because we only need to describe how r relates to z.
$$
z = mr
$$
As long as this equation holds, points \((r, \theta, z)\) with any angle \(\theta\) are on the double cone.

Spherical

Spherical coordinates are a triple \((\rho, \phi, \theta)\), where \(\rho\) is the Euclidean distance of the (Cartesian) point \(P = (x, y, z)\) from the origin, \(\phi\) is the angle between the positive z-axis and the line through the origin and point \(P\) (in the graph above \(\phi = \pi / 2 - \alpha\), in general \(\tan \phi = \sqrt{x^2 + y^2} / z\)), and \(\theta\) is the angle of the polar coordinates, as in the cylindrical coordinates.
We already saw that we don't have any constraint on the angle \(\theta\) and since the double cone is infinite, we also don't have a constraint on the Euclidean distance.
So we can drop another variable from our equation and end up with the spherical equation using only a single coordinate.
$$
\tan \phi = 1/m
$$