Dandelin Spheres: The Two-Sphere Proof That a Cone Slice Is an Ellipse

The whole proof rests on one lemma

Take a point P outside a sphere of centre O and radius r, and draw any line through P that just touches the sphere, at a point T. The radius OT is perpendicular to a tangent line, so OTP is a right triangle and |PT| = √(|PO|² − r²). That expression does not mention T at all. Every tangent segment from P to the sphere has the same length, and the tangent points sweep out a circle around the line PO.

This is the three-dimensional twin of a fact from plane geometry that is roughly 2,300 years old — the two tangents from an external point to a circle are equal (Euclid, Elements, Book III). Dandelin's argument uses nothing beyond it: no coordinates, no quadratic form, no discriminant, no calculus.

One consequence deserves its own sentence, because the construction leans on it hard. The apex of the cone is an external point for every sphere inscribed in that cone, so the tangent length from the apex to a given inscribed sphere is one fixed number. That is exactly what makes “the slant distance from the apex to the circle of contact” a well-defined quantity rather than a direction-dependent one.

Inflating the spheres — and proving they exist

Put the apex at the origin with the axis along z and half-angle α, so the cone is x² + y² = (z·tan α)². A sphere centred on the axis at height h and inscribed in that cone has radius r = h·sin α, and it touches the cone along a horizontal circle at slant distance L = h·cos α from the apex.

Now cut with a plane, written z = x·tan β + c, where β is its tilt away from horizontal. The distance from the sphere centre (0, 0, h) to that plane is |c − h|·cos β. Tangency means that distance equals the radius, |c − h|·cos β = h·sin α, and for an ellipse (0 ≤ β < 90° − α) that equation has exactly two positive roots:

h₁ = c·cos β / (cos β + sin α)  and  h₂ = c·cos β / (cos β − sin α),

one centre below the plane and one above. The animation shows the spheres swelling until they jam; the algebra above, or equivalently the intermediate value theorem applied to the continuous function h ↦ |c − h|·cos β − h·sin α, is what actually guarantees that they exist. Drawing a sphere is not the same as showing one fits.

Look at where h₂ blows up: at cos β = sin α, that is β = 90° − α. That is precisely the parabola, and the formula is telling you the truth — the upper sphere has run away to infinity, which is why a parabola gets only one focus.

The proof, in four lines

Call the two spheres S₁ (below the plane) and S₂ (above), let F₁ and F₂ be the points where they touch the cutting plane, and let their circles of contact with the cone sit at slant distances L₁ < L₂ from the apex. Take any point P on the section and draw the generator — the straight ruling of the cone — from the apex through P. It crosses the lower contact circle at T₁ and the upper one at T₂.

• PF₁ and PT₁ are both tangent segments from P to S₁: the cutting plane touches S₁ at F₁ and P lies in that plane, while the generator touches S₁ at T₁. So PF₁ = PT₁.
• By the same argument on S₂, PF₂ = PT₂.
• P sits between the two contact circles on its own generator, so PT₁ + PT₂ = T₁T₂ = L₂ − L₁.
• Therefore PF₁ + PF₂ = L₂ − L₁, a number built only from the cone and the plane.

Nothing in those four lines mentions where P was. Every point of the section has the same sum of distances to F₁ and F₂ — the two-pins-and-a-string definition of an ellipse — with major axis 2a = L₂ − L₁. The contact circles never move, so the constant cannot move either.

A worked cone, with all the numbers

Take a cone of half-angle α = 30°, so x² + y² = z²/3, and cut it with the plane z = x/√3 + 2, tilted β = 30°. Everything below is exact, not rounded-to-fit.

The section. Eliminating z gives (x − √3/4)² + (9/8)y² = 27/16 for the shadow of the curve on the ground. Un-squashing by the factor 1/cos 30° along the tilt direction, the curve in its own plane has semi-major axis a = 3/2, semi-minor axis b = √(3/2) ≈ 1.22474, linear eccentricity c = √3/2 ≈ 0.86603 and eccentricity e = 1/√3 ≈ 0.57735.

The spheres. The tangency equation gives centres at z = 3 − √3 ≈ 1.26795 and z = 3 + √3 ≈ 4.73205, with radii r₁ = (3 − √3)/2 ≈ 0.63397 and r₂ = (3 + √3)/2 ≈ 2.36603. They touch the cone along circles at slant distances L₁ = (3√3 − 3)/2 ≈ 1.09808 and L₂ = (3√3 + 3)/2 ≈ 4.09808.

The check. L₂ − L₁ = 3 = 2a, exactly. The contact points with the plane come out at F₁ = (−0.31699, 0, 1.81699) and F₂ = (1.18301, 0, 2.68301), a distance √3 apart — and 2ae = 3·(1/√3) = √3. Sampling any point of the curve gives PF₁ + PF₂ = 3.000000 to every decimal place a computer will print.

A bonus identity. Here r₁r₂ = (3 − √3)(3 + √3)/4 = 6/4 = 3/2 = b². That is not a coincidence of this example: in general both sides equal c²·cos²β·sin²α / (cos²β − sin²α), so the semi-minor axis of the section is always the geometric mean of the two Dandelin radii, b = √(r₁r₂).

Hyperbola, parabola, and where eccentricity comes from

Hyperbola. Tilt the plane past β = 90° − α and it cuts both nappes of the double cone. Now one sphere fits in each nappe. For P on one branch, T₁ lies on the contact circle in P's own nappe and T₂ on the circle in the opposite one, so the two tangent lengths sit on opposite sides of the apex: PT₂ − PT₁ = L₁ + L₂. The same equal-tangent lemma then gives |PF₂ − PF₁| = L₁ + L₂, constant — the difference definition of a hyperbola, with 2a = L₁ + L₂.

Parabola. At β = 90° − α the plane is parallel to a generator and only one sphere fits. Adolphe Quetelet supplied the missing half. Let H be the horizontal plane containing that sphere's contact circle, and let d be the line where H meets the cutting plane. For P on the section, the distance from P to the plane H is cos α·PT, and the in-plane distance from P to the line d is that same height divided by sin β. Dividing,

PF / dist(P, d) = sin β / cos α.

Eccentricity, for free. That derivation never assumed a parabola, so it holds for every section: e = sin β / cos α, where β is the tilt of the cutting plane from the plane perpendicular to the axis and α the cone's half-angle. Flat plane, β = 0: e = 0, a circle. Plane parallel to a generator, β = 90° − α: e = cos α / cos α = 1, a parabola. Steeper still: e > 1, a hyperbola. In the worked cone above, e = sin 30° / cos 30° = 1/√3 ≈ 0.5774, matching the axes; in the animation on this page the cone has α = 20° and β = 35°, giving e = 0.5736 / 0.9397 ≈ 0.6104.

What the picture proves, and what it quietly assumes

Dandelin's diagram is a genuine proof, not an illustration — but only once three things are said out loud.

1. The cone must be a cone of revolution. The tangent lines from a point to a sphere form a right circular cone, so a sphere can only be inscribed in — touching along a curve of — a right circular cone. Oblique circular cones still cut conics out of planes, but you reach that by a projective or algebraic route, not by this one.

2. The plane must miss the apex. A plane through the apex gives a point, a single line, or two crossing lines, and no inscribed sphere is usefully tangent to it. Those degenerate cases are genuinely outside the theorem, not special cases of it.

3. Containment is not yet equality. The four-line argument shows every point of the section lies on the ellipse with foci F₁, F₂ and constant 2a. To conclude the section is that ellipse you need one more step: the section is a compact, connected simple closed curve sitting inside another simple closed curve, and a subset of a topological circle that is itself a topological circle must be the whole circle. Most textbook presentations skip this line; it is one sentence, and it is the difference between “satisfies the equation” and “is the curve”.

One thing the construction does not prove is the converse — that every ellipse arises as a cone section. That direction is easy but separate: given any e in (0, 1), pick any half-angle α, solve sin β = e·cos α for the tilt (possible because e·cos α < 1), then scale c until 2a comes out right.

Dandelin, Quetelet, and 2,000 years of conics

Apollonius of Perga (c. 262–190 BC) wrote the eight-book Conics that gave the curves their names — ellipse, parabola, hyperbola — and proved the constant-sum and constant-difference focal properties in Book III. He got there by the Greek method of “application of areas”, with no spheres anywhere, and he never used a directrix. The focus–directrix property is first recorded by Pappus of Alexandria around AD 320 in Book VII of his Collection, where it is usually taken to preserve a lost result of Euclid's.

The sphere argument arrived only in the nineteenth century. Germinal Pierre Dandelin (1794–1847) — a Belgian mathematician and military engineer who taught at Liège — published it in the Nouveaux Mémoires of the Royal Academy of Brussels in 1822. His Brussels colleague Adolphe Quetelet (1796–1874), better remembered today for founding social statistics and for the Quetelet index that became BMI, added the directrix half, which is why the construction is sometimes called the Dandelin–Quetelet spheres. Independently, Pierce Morton published a sphere proof of the focus–directrix property in the Transactions of the Cambridge Philosophical Society in 1829.

Dandelin's name survives in a second, unrelated place: the Dandelin–Gräffe method for finding polynomial roots by repeatedly forming the polynomial whose roots are the squares of the originals, separating them geometrically. Dandelin described it in 1826, Lobachevsky in 1834, and Gräffe in 1837.

The reason the two spheres still get taught is pedagogical rather than historical. Almost every course defines a conic twice — once by slicing a cone, once as the locus of points with a fixed focal sum, difference, or ratio — and then quietly never connects the two. Dandelin's construction is that connection, and it fits on a single page with one lemma and no algebra.

What Dandelin's construction produces for each kind of plane section

Section	Spheres that fit	What the contact points give	Tilt and eccentricity
Circle	2 — one above the plane, one below	both touch the plane at the same point: the centre, a doubled focus	tilt β = 0°, e = 0
Ellipse	2 — both inside the same nappe	the two foci, with PF₁ + PF₂ = 2a = L₂ − L₁	0° < β < 90° − α, so 0 < e < 1
Parabola	1 — the second would have to sit at infinity	the single focus; the contact circle's plane meets the cutting plane in the directrix	β = 90° − α exactly, e = 1
Hyperbola	2 — one in each nappe	the two foci, with |PF₂ − PF₁| = 2a = L₁ + L₂ measured through the apex	β > 90° − α, so e > 1
Degenerate (plane through the apex)	none usable — a plane through the apex either misses every inscribed sphere, touches them all along the grazing line, or cuts straight through them	a point, a single line, or two crossing lines — no foci at all	any β; the construction simply breaks