# Definition of an ellipsoid based on its focal points

I have a question concerning the formulation of an (3D) ellipsoid. The most common definition for an ellipsoid seems to be:

$E = \{ x=\left( x_1, \dots x_n \right)^T \in R^n: \sum_{i=1}^n \left( \frac{x_i}{r_i} \right)^2 = 1 \}$

based on the different radii $r_i$. However for an ellipse (2D) I also found the following definition:

$E = \{ x \in R^2: \| x - f_1 \| + \| x - f_2 \| = c \}$

which is based on the two focal points $f_1$ and $f_2$ of an ellipse. Now my question is, if there exist such kind of definitions also for ellipsoid in 3D or higher? How many focal points can an ellipsoid in 3D or higher actually have? I am especially interested in ellipsoids where all radii $r_1, \dots, r_n$ can be different (like a Scalene or triaxial ellipsoids).

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We should investigate if any biaxial (that is, by a plane containing any two ellipsoid axes) section of an ellipsoid gives an ellipse. Both answers give strange results, like having ellipse with four foci or with no foci at all. – mbaitoff Feb 1 '11 at 11:17
If I remember correctly, the analogue of the pair of focal points for an ellipsoid in 3D are a pair of curves, namely an ellipse and a hyperbola (in two orthogonal planes). Unless someone else gives an answer, I will try to look into this and return later. – Hans Lundmark Feb 1 '11 at 11:20
Hmm, maybe this is it: books.google.com/… – Hans Lundmark Feb 1 '11 at 11:22

$$\frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}+\frac{z^2}{c^2+s}=1$$

where $a>b>c\geq 0$, there're one focal ellipse namely,

$$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-c^2}+\frac{y^2}{b^2-c^2} &=& 1 \\ z &=& 0 \end{array} \right.$$

and one focal hyperbola namely,

$$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-b^2}-\frac{z^2}{b^2-c^2} &=& 1 \\ y &=& 0 \end{array} \right.$$

For confocal paraboloids,

$$\frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}=z+\frac{s}{4}$$

where $a>b\geq 0$, there're two focal parabolae namely,

$$\left \{ \begin{array}{rcl} \displaystyle \frac{x^2}{a^2-b^2} &=& \displaystyle z-\frac{b^2}{4} \\ y &=& 0 \end{array} \right.$$

and

$$\left \{ \begin{array}{rcl} \displaystyle \frac{y^2}{b^2-a^2} &=& \displaystyle z-\frac{a^2}{4} \\ x &=& 0 \end{array} \right.$$

Note that the two focal conics are called conjugate conics and their eccentricities are reciprocal to each other.

Any point on the conjugate conics is the radiant point of the enveloping cone for that conic and its Dandelin sphere.

P.S.:

The lines of curvature of an ellipsoid actually are the geodesic ellipses and geodesic hyperbolae from two adjacent umbilical points.

See more details here.

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Hint:

By symmetry, a quadric with two focal points is an ellipsoid of revolution and can't be more.

In 2D, a conic is defined by $5$ independent parameters, which can be the coordinates of two foci and the sum of distances.

A general quadric needs $9$ parameters, which cannot equal $3n+1$. (Not a proof but a serious clue; similarly, the reduced conic has only $3$ parameters, which cannot equal $3n-5$, where $6$ DOFs have been absorbed by a similarity transform.)

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