Two balls wedged into a sliced cone prove where an ellipse's foci are.
| PF₁ = P to sphere 1 | — |
| PF₂ = P to sphere 2 | — |
| PF₁ + PF₂ | — |
Where's the maths? This is geometry — the geometry of circles and solids you already meet in Sec 2 and Sec 3. Slice a cone with a flat plane and the rim of the cut is an ellipse, an oval with two special points inside it called the foci. The two balls in the picture have been squeezed into the cone, one from above the plane and one from below, until each one touches the cone all the way round and just kisses the cutting plane at a single point. Those two kissing points are exactly the foci — and that is not a coincidence, it is forced by one small circle fact you learn in class: two tangents drawn from an outside point to a circle (or a sphere) are equal in length. Follow the straight line of the cone through your point P and the same two lengths turn up twice, once measured to the plane and once measured along the cone. Adding them gives the distance between the two circles where the balls touch the cone — a number that has nothing at all to do with where P sits. That is why PF₁ + PF₂ never changes, and it is the whole proof, found by the Belgian mathematician Germinal Dandelin in 1822.
Beyond the syllabus: the ellipse itself is not an O-Level curve.
An ellipse can be defined without any cone at all: pin two points F₁ and F₂, take a loop of string, and trace every point P whose two distances add to the same total. That total is written 2a, where a is the ellipse's semi-major axis. The question Dandelin answered is: why does slicing a cone give one of those curves, and where exactly are the pins?
Drop a ball into the cone above the plane and let it settle. It grows until it is stuck: it touches the cone in a complete circle and touches the plane at one point, F₁. Do the same from the other side and you get the second ball and F₂. Neither ball was placed by eye — the computer solves for the one centre on the axis whose distance to the plane equals its distance to the cone.
Now take any point P on the cut curve and draw the straight line of the cone through P (its generator). From P there are two tangents to the first ball: the segment PF₁ lying in the plane, and the segment PT₁ running along the generator to the contact circle. Tangents from an external point are equal, so PF₁ = PT₁. The same argument gives PF₂ = PT₂.
Add them: PF₁ + PF₂ = PT₁ + PT₂ = T₁T₂, the distance along the generator between the two contact circles. Every generator meets those two circles the same distance apart, so this is a fixed number — it equals 2a. Move P wherever you like: the sum in the readout does not budge, so F₁ and F₂ really are the foci.
Tilt the plane further and the far ball has to grow to keep touching. At the moment the plane becomes parallel to a generator it escapes to infinity: the curve is a parabola and only one Dandelin sphere survives. Tilt past that and the second ball reappears inside the other half of the double cone; now the same argument gives a constant difference |PF₁ − PF₂|, and the curve is a hyperbola.
Controls: drag to orbit, scroll or pinch to zoom, drag the point P straight along the curve, or use the sliders. Look along the axis aims the camera down the cone so the cut looks like a flat ellipse.