As a few people pointed out in replies, here and elsewhere, we can get the same result using a fixed linear map T from the unit n-sphere to the ellipsoid. T will map any bounding n-cube of the n-sphere to a suitable parallelotope of volume 2^n det(T).

Almost. You need an orthogonal projection to map the square to parallelograms, whereas a perspective view will send parallel sides of the square to lines that converge on vanishing points.

Since those 𝑥ᵗ will be orthogonal to norm(𝑥ˢ), each face will be tangent to the ellipsoid at its centre.

So det(𝑋) will always be the same, the product of all the semi-axes, and the volume of the parallelotope that encloses the ellipsoid will be 2ⁿ det(𝑋), with each of its faces formed by taking ±𝑥ˢ as the centre and adding ±𝑥ᵗ, for all the choices of t≠s.

If we define the matrix 𝑋 to have the vectors 𝑥ˢ as its rows, and the matrix 𝐴 to have diagonal elements 1/𝑎ᵢ² and zeroes elsewhere, then the dot products above correspond to:

𝑋 𝐴 𝑋ᵀ = 𝐼

If we take the determinants of both sides of this equation, we have:

det(𝑋)² det(𝐴) = 1

Suppose we choose n points on the surface of the ellipsoid, 𝑥ˢ, s=1,...n, such that:

𝑥ᵗ·norm(𝑥ˢ) = 0 when t≠s

We will also have:

𝑥ˢ·norm(𝑥ˢ) = ∑ᵢ₌₁ⁿ𝑥ˢᵢ²/𝑎ᵢ² = 1

A vector normal to the surface can be found from the gradient of the left-hand side of this equation; since that function is constant on the surface, the gradient is orthogonal to the surface.

norm(𝑥) = (𝑥ᵢ/𝑎ᵢ²)

It’s not hard to prove that this generalises to n dimensions.

Suppose we have an n-ellipsoid given by the equation:

∑ᵢ₌₁ⁿ𝑥ᵢ²/𝑎ᵢ² = 1

where the 𝑥ᵢ are coordinates (x,y,z,...) and the 𝑎ᵢ are the semi-axes in each direction.

Every parallelepiped that you place around an ellipsoid whose faces are tangent to the ellipsoid at their centres has the same volume for a given ellipsoid: 8 a b c, where a, b and c are the semi-axes of the ellipsoid.

You made me Google it.

www.researchgate.net/publication/...

This is a geometrical way of saying that putting tension on the string won’t make it slide along these curves, because the sum of the distances of the adjoining segments is a local minimum, so it can’t offer up slack by moving the point.

The portions of the string where it departs from the hyperbola or the ellipse both lie on cones whose axis is a tangent to the curve at that point, and which make an angle with the tangent that is the same as the adjacent segment of the string.

Here is a version where the point on the ellipse is held still while the point on the hyperbola is swept along it.

This construction was found in the late 1800s by the German mathematician Staude (Wikipedia says the famous physicist Maxwell also made a start on the problem a bit earlier). I haven’t read Staude’s proof, but you can read a modern treatment here:

arxiv.org/abs/1707.012...

Pin one end of a string to one focus of the hyperbola, and the other to the opposite focus of the ellipse, and then pull the string taut, while also requiring it to pass through whatever point on each curve minimises the distance. The point the string reaches in between will trace out an ellipsoid!