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Why slicing a cone gives an ellipse (beautiful proof)

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Summary

The video presents a geometric proof that slicing a cone with a plane produces an ellipse with the same defining property as the curve drawn using two thumbtacks and a taut string.

Three ways to describe an ellipse

  • A stretched circle: Stretching a circle in one direction produces an ellipse.
  • The two-focus definition: An ellipse is the set of points for which the sum of the distances to two fixed points—the foci—is constant. This is the property behind the thumbtack-and-string construction.
  • A conic section: An ellipse is formed where a plane cuts through a cone at an angle shallower than the cone’s slope.

These constructions may look unrelated, but the proof shows why the cone’s intersection has the constant-distance-sum property.

Eccentricity

  • Eccentricity describes how elongated or “squished” an ellipse is.
  • A circle has eccentricity 0; as an ellipse becomes more elongated, its eccentricity approaches 1.
  • In the two-focus definition, eccentricity depends on the distance between the foci relative to the ellipse’s longest axis.
  • For a cone section, eccentricity depends on the angle of the slicing plane.

The Dandelin-sphere proof

  1. Add two spheres. Place one sphere above the slicing plane and another below it. Each sphere is tangent to the cone along a circle and tangent to the plane at one point.
  2. Treat the plane-tangency points as candidate foci. These are the two points from which distances to a point on the intersection curve will be measured.
  3. Choose a point (q) on the ellipse. Through (q), consider the line running along the cone between the two circles where the spheres touch it. This line is split at (q) into two segments.
  4. Use equal tangent lengths. From a point outside a sphere, any two tangent segments to that sphere have equal lengths:
    • For the upper sphere, the segment from (q) to its cone-tangency circle equals the segment from (q) to the upper sphere’s plane-tangency point.
    • For the lower sphere, the segment from (q) to its cone-tangency circle equals the segment from (q) to the lower sphere’s plane-tangency point.
  5. Add the two equalities. The distances from (q) to the two proposed foci therefore add up to the length of the cone segment between the two circles of tangency.
  6. Recognize that the sum is constant. That cone segment has the same length wherever (q) lies on the intersection curve. Thus, every point on the curve has the same sum of distances to the two proposed foci.

The curve therefore satisfies the two-focus definition of an ellipse. The spheres used in the proof are called Dandelin spheres.

Further implications and lessons

  • The same idea can show that slicing a cylinder at an angle produces an ellipse.
  • The speaker notes that projecting a shape from one plane onto another tilted plane stretches it, connecting the stretched-circle definition to the other descriptions.
  • The proof illustrates that mathematics often has multiple useful definitions of the same object; showing that those definitions are equivalent can matter more than choosing one as uniquely fundamental.
  • The key creative step is introducing the two spheres. The speaker argues that such ideas need not be mysterious acts of genius: practice with geometry makes tools such as circles, spheres, and length comparisons more natural to reach for.

Speakers and sources featured

  • Narrator: The video’s presenter, explaining the definitions, proof, and reflections on mathematical creativity.
  • Paul Lockhart: His book Measurement is credited by the narrator as the source where the proof was first encountered; the video also refers to Lockhart’s remarks about the proof.
  • E. H. Dandelin: Credited with the proof in 1822; the spheres in the argument are named after him.
  • A Reddit user: Mentioned as having asked why the thumbtack and cone-slicing definitions of an ellipse agree.
  • Feynman’s Lost Lecture: Referenced as context for the narrator’s earlier video about planetary orbits; it is not presented as a speaker in this proof.

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