Degenerate conic sections

Picture a double-napped right circular cone: two cones joined tip to tip at a single point, the vertex $V$, with an axis running through $V$ and generators (the slant lines on the surface) making a fixed half-angle $\alpha$ with that axis. For every non-degenerate conic — circle, ellipse, parabola, hyperbola — the cutting plane misses the vertex entirely; it slices cleanly through the body of the cone. The classification then depends only on the angle $\beta$ the plane makes with the axis, compared to $\alpha$:

• $\beta = 90^\circ$ gives a circle,
• $\alpha < \beta < 90^\circ$ gives an ellipse,
• $\beta = \alpha$ gives a parabola,
• $\beta < \alpha$ gives a hyperbola.

Degenerate conics are the same three angle regimes, but with one extra condition: the plane now passes through the vertex $V$. Once it touches the tip, the smooth curve collapses:

• A plane through $V$ that behaves like the circle/ellipse case (roughly perpendicular to the axis, or at any angle steeper than a generator) meets the cone only at the single point $V$ itself — the section is a point.
• A plane through $V$ that behaves like the parabola case ($\beta = \alpha$) contains exactly one generator of the cone — the section is a single straight line.
• A plane through $V$ that behaves like the hyperbola case ($\beta < \alpha$) cuts into both nappes at once, and its intersection with the cone is exactly two generators crossing at $V$ — the section is a pair of intersecting straight lines.

where: $\alpha$ is the cone's fixed half-angle between the axis and any generator, $\beta$ is the angle the cutting plane makes with the axis, and $V$ is the vertex — the single point common to both nappes. No coordinates or equations are needed here; the whole classification is a picture-reading exercise about where the plane sits relative to $V$, $\alpha$, and $\beta$.

The reason this collapse happens is purely about what the plane can meet once it is forced to pass through $V$. Away from the vertex, a cutting plane sweeps through the full body of the cone and traces out a continuous curve because the cone's surface bulges outward on both sides of the plane. At the vertex itself, the two nappes meet at nothing but a single point — there is no "body" left to slice, only the tip where all the generators converge. So when the plane is anchored at $V$, everything depends on how many generators that plane happens to contain. A generic plane through $V$ (one that is not parallel to any generator) touches the cone's surface only at $V$, because every generator leaves $V$ in a different direction and the plane can only intersect a line at one point unless it actually contains that line. That is why the ellipse-type and circle-type planes through the vertex shrink to a single point: they carry no generator, so they meet the cone precisely at $V$ and nowhere else. A plane through $V$ that is tilted to exactly match one generator's direction does contain that whole line, so the intersection is that one generator — a straight line, the degenerate parabola. A plane steep enough to cut both nappes (the hyperbola regime, $\beta < \alpha$) contains two generators, one on each nappe, both passing through $V$; the intersection is those two lines crossing at the vertex. This is exactly the logic used to classify the non-degenerate conics by comparing $\beta$ to $\alpha$ — degenerate conics use the identical angle test, just applied at the one point where the cone pinches to nothing.

Problem. A plane through the vertex of a cone (half-angle $\alpha = 35^\circ$) makes an angle $\beta = 20^\circ$ with the axis. Identify the degenerate conic and justify the shape from the angle comparison.

Step 1. Compare the given angle to the cone's half-angle: $\beta = 20^\circ$ and $\alpha = 35^\circ$, so $\beta < \alpha$.

Step 2. Recall the non-degenerate rule: $\beta < \alpha$ is the hyperbola regime, since a plane that shallow (away from the vertex) would cut both nappes and produce two separate branches.

Step 3. Because this plane passes through $V$ rather than avoiding it, the "cutting both nappes" behaviour collapses onto the vertex itself: the plane meets each nappe along one generator, and both generators pass through $V$.

Answer: the section is a pair of intersecting straight lines, crossing at the vertex $V$ — the degenerate form of a hyperbola.

Competitive stretch

If instead $\beta = \alpha = 35^\circ$ for a plane through the same vertex, the plane matches the direction of exactly one generator, so the section degenerates to a single straight line (the degenerate parabola) rather than two lines — a one-degree shift in $\beta$ changes the answer from two crossing lines to one line, which is why the angle comparison must be checked precisely, not just estimated as "small."

• Treating degenerate conics as a fourth family unrelated to the circle/ellipse/parabola/hyperbola — they are the same classification, just with the plane pinned to the vertex.
• Forgetting that passing through the vertex is the defining condition — angle alone never makes a conic degenerate; the plane must actually touch $V$.
• Assuming any plane through the vertex that is not along a generator gives a line — it gives a point unless it exactly contains a generator ($\beta = \alpha$) or cuts both nappes ($\beta < \alpha$).
• Mixing up which shape goes with which regime: point for $\beta \ge \alpha$ (ellipse/circle-type), one line for $\beta = \alpha$ (parabola-type), two lines for $\beta < \alpha$ (hyperbola-type) — always through $V$.

Ask the tutor: Want to see why an ellipse "shrinking" toward the vertex jumps straight to a point instead of getting smaller and smaller — try asking for the limiting-plane picture worked through step by step.