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A conic section is obtained by intersecting a plane with a double circular cone. Cuts away from the vertex produce an ellipse, a parabola or a hyperbola; a perpendicular cut is the circle case of an ellipse. A cut through the vertex can instead produce a degenerate conic. One simple surface therefore connects several familiar plane curves.^[ref]

Consider the original familySubstituting the plane equation into the double circular cone equation givesThus $m=0$ gives a circle, $m=3/5$ an ellipse, $m=1$ a parabola, and $m=7/5$ a hyperbola. The last cut meets both sheets of the double circular cone, producing its two branches. These parameters specify a mathematical family, rather than measurements of a physical object.

The Cartesian coordinate system descriptions of these curves are second-degree equationsA rotation and a translation expose their principal directions and offsets. The conic type discriminant identifies the ordinary nondegenerate real type, while a separate degenerate conic check prevents confusing two crossing straight lines with a hyperbola.

This is useful because a difficult-looking equation can be understood through a small set of geometric features: foci of a conic, directrices of a conic, eccentricity of a conic and the semi-major and semi-minor axes. The same ellipse can be traced by a constant-length string or parametrized by $x=a\cos t$, $y=b\sin t$. These are different descriptions of one curve, with different advantages for construction and calculation.

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.

Put $v=x-a$ and decompose it into axial and perpendicular parts, $v=(v\cdot n)n+v_{\perp }$. For the nondegenerate semi-angle $0<\alpha <\pi /2$, the equation is equivalent toThus each plane perpendicular to $n$ at axial distance $s$ from $a$ meets the surface in a circle of radius $|s|\tan \alpha$. Both signs of $s$ are allowed, and $s=0$ gives the vertex. This proves that it is a double circular cone with the specified vertex and axis; $\alpha$ is the angle each generator makes with either direction of the axis.

For the intersection calculation, define the symmetric matrix $Q={\cos }^2\alpha I-n{n}^T$. The two equations are $(x-a_i{)}^{T}Q(x-a_i)=0$. Their difference givesWith $b=a_1-a_2$ and $m=(a_1+a_2)/2$, symmetry of $Q$ makes the last two terms $2m\cdot Qb$. Hence the intersection plane of parallel congruent double cones isThe unnormalized normal vector is $v=Qb=b{\cos }^2\alpha -n(n\cdot b)$. Expanding its squared norm givesEquivalently, if $b=b_{\perp }+(n\cdot b)n$, this is ${\cos }^4\alpha |b_{\perp }{|}^2+{\sin }^4\alpha (n\cdot b{)}^2>0$. Thus normalization is legitimate and yieldsThe assumption of a genuine cone matters: at the degenerate limits $\alpha =0$ or $\pi /2$, the surface collapses respectively to an axis or a plane and this normal need not exist.

For $\mathbf{r}\ne 0$, the dot product equation says that the angle $\theta$ between $\mathbf{r}$ and the oriented unit vector $\mathbf{u}$ satisfies $\cos \theta =\alpha$. Thus $\theta =\arccos \alpha$ is fixed. More explicitly, using the orthogonal projection from the introductory solution,Consequently the surface is one infinite nappe of a right circular cone, with apex at the origin, including the apex itself. The cone's axis points along $\mathrm{sgn}(\alpha )\mathbf{u}$, and its acute semi-angle is $\arccos |\alpha |$. Equivalently, $\alpha$ is the cosine of the angle from the oriented $\mathbf{u}$ axis. Squaring the original equation without retaining the sign of $s$ would incorrectly add the opposite nappe of a double circular cone. No base disk is included in this infinite lateral surface.

Away from the vertex, use the defining function $H=z^2-x^2-y^2$. Its gradient is normal to the smooth part of the double circular cone, andon the surface. Hence the field is tangent at every regular point of the cone. At the origin the cone has no unique tangent plane, but $\mathbf{F}(0)=0$, which belongs to its tangent cone. An equivalent argument, valid also at the vertex, uses the integral curves of a vector field: $\dot{\mathbf{x}}=\mathbf{x}$ has solution $\mathbf{x}(t)=e^t\mathbf{x}(0)$, and these dilations preserve $H=0$. This is an example of radial dilation tangency to homogeneous zero sets.

At every regular point of the double circular cone, its normal is perpendicular to $\mathbf{F}$ by part (a). Therefore the integrand of the flux integral vanishes pointwise:The vertex, if included, is a single point of zero surface area and does not alter the surface integral. The answer is independent of both the subset and its orientation. Tangency, rather than a cancellation between different parts of the surface, explains the zero flux.

Close the truncated two-sheet double circular cone by the lateral circular cylinder $r=1$, $-1\le z\le 1$. Together these form the boundary of the solid in part (c); no planar caps are required because the cone meets the circular cylinder along both end circles. On the circular cylinder the outward unit normal is $(\cos \theta ,\sin \theta ,0)$, so $\mathbf{F}\cdot \mathbf{n}=1$, and $dS=d\theta dz$. Its flux integral isThe divergence theorem now gives ${\int }_S\mathbf{F}\cdot d\mathbf{S}+4\pi =4\pi$, using the volume integral from part (c). Henceagreeing with tangency. The isolated vertex can be handled by excising a ball of radius $\epsilon$ and passing to the limit: since $|\mathbf{F}|=O(\epsilon )$ there and the added area is $O({\epsilon }^2)$, its extra flux is $O({\epsilon }^3)$ and tends to zero.