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Given an infinite double cone of the form $ax^2+by^2+cz^2+dxy+exz+fyz=0$, how can I get the slope of the double cone, the radius at a given height along the cone's axis, and the angles of rotation?

For example:

Given $2x^2+2y^2-0.5z^2=0$ (visualization):

  • the radius at the height $-1$ is $0.5$
  • the slope is $\pm2$
  • the angles of rotation are $0$ and $0$ (unrotated)

Note that the radius-at-height and slope attributes must be as if the cone were not rotated. For example:

Given $2x^2+2y^2-0.5z^2-xz=0$ (visualization):

  • the radius at the height $-1$ is $0.5$
  • the slope is $\pm2$
  • the angles of rotation are $0$ and $\sim\frac{\pi}{6}$ (the second is visually observed)

How can I solve for the various attributes algebraically?

Dr. Vortex
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    You have arranged it so the singular point is the origin, so if the cone is circular you can match the equation with a multiple of $(\cos{\theta})^2(x^2+y^2+z^2)-(n_x x+ n_y y+n_z z)^2=0,$ and determine $n_1,n_2,n_3,\theta$ where $[n_1,n_2,n_3]$ is a unit vector along the axis and $\theta$ is the apex angle. – Jan-Magnus Økland Feb 23 '23 at 17:38
  • @Jan-MagnusØkland Would you mind writing an answer? – Dr. Vortex Feb 23 '23 at 19:15
  • The operative word being if. As we've seen in another of your questions, any conic section can be cut out by a plane and a right circular cone, but when you start with a general cone with cone point at the origin, the cone is not likely to be circular, as you can see by a parameter count. – Jan-Magnus Økland Feb 24 '23 at 07:23

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