Double circular cone ($z^2=x^2+y^2$)

The union of the two infinite lateral conical sheets meeting at a common vertex is a double circular cone. The displayed example has axis $z$ and semi-angle $\pi /4$. It is smooth away from the vertex, where its defining gradient is nonzero. A right circular cone truncated on one side is a related bounded solid; this surface includes both axial directions and has no base disks.

Let two double circular cones have vertices $a_1,a_2$, common unit axis $n$, and semi-angle $0<\alpha <\pi /2$. Put $Q={\cos }^2\alpha I-n{n}^T$ and $b=a_1-a_2\ne 0$. Their equations are $(x-a_i{)}^{T}Q(x-a_i)=0$. Subtracting cancels the common quadratic term, so every intersection point satisfiesSince $Q$ has eigenvalues ${\cos }^2\alpha$ perpendicular to $n$ and $-{\sin }^2\alpha$ along $n$, $Qb\ne 0$. Thus the intersection is contained in a plane, possibly with an empty intersection. Its unit normal is $Qb/|Qb|$, with $|Qb{|}^2=|b{|}^2{\cos }^4\alpha +(n\cdot b{)}^2(1-2{\cos }^2\alpha )$. Degenerate endpoint semi-angles can make the normal vanish and must be treated separately.