The Journal of Online Mathematics and Its Applications, Volume 8 (2008)

The Most Marvelous Theorem in Mathematics, Dan Kalman

As an alternate definition of an ellipse, we begin with two fixed points in the plane. Now consider any point whose distances from these two points add up to a fixed constant `d`. The set of all such points is an ellipse. The two fixed points that were chosen at the start are called the `foci `(pronounced foe-sigh) of the ellipse; individually, each of these points is called a `focus `(pronounced in the usual way).

For example, if you specify points `A` and `B` in the plane, 5 units apart, then look for every possible point `C` whose distances from `A` and `B` add up to 7, these points form an ellipse. One way to imagine drawing the ellipse in this example is to cut a piece of string, 7 units long, and tack the ends to the points `A` and `B`. Stretch the string taut with the point of your pencil, and put the pencil on the paper. From the pencil point to `A` and `B` the distances must total 7 units, since together these lengths use up the entire string. This shows that your pencil point is on the ellipse. Now move the pencil, keeping the string taut, and you will trace the ellipse. This idea can be visualized in the animated graphic below, which is located at Wolfram MathWorld. As the point moves around the ellipse, the two red line segments always combine to the same total length.

The pencil and string construction in the graphic would not be possible with a string less than 5 units long. In general, we need a string longer than the distance between `A` and `B`. In the extreme case, with a string exactly long enough to reach from `A` to `B`, the string becomes a straight line segment. Then the pencil slides along the taut string, tracing the line segment. This corresponds to a degenerate ellipse. To avoid this situation we insist that `d` is greater than the distance between the given points `A` and `B`.

The fact that every ellipse can be defined using two foci and a specified number `d` will be used frequently in the discussions to follow. For reference, we will call this the Two Focus Property of an ellipse. It has the following official statement:

For every ellipse `E` there are two distinguished points, called the foci, and a fixed positive constant `d` greater than the distance between the foci, so that from any point of the ellipse, the sum of the distances to the two foci equals `d`.

The goal for this page is to show that the Two Focus Property is a valid alternate definition of an ellipse. That means we have to show that every ellipse has the two focus property, and that any curve with the two focus property is an ellipse. The details are a bit involved, so here is a preview. As a first step, we show that every ellipse is made up of exactly the points defined by the Two Focus Property for a suitable choice of the foci and the constant `d`. That in itself requires two parts: showing that points of the ellipse satisfy the Two Focus Property, and also showing that no other points do. Then we still have to consider the converse. That means showing that the set of points defined by the Two Focus Property for a particular choice of the foci and `d` is necessarily an ellipse. That will complete the overall demonstration.

To get started, let us see that a curve given by the standard ellipse equation

(1) |

always satisfies the Two Focus Property. There are three possible cases, depending on whether `a` is equal to, greater than, or less than `b`, and we will consider each case separately.

Here the curve is a circle and we take both foci to be at the center of the circle. From any point on the circle, the distance to the center is a constant, so the sum of distances to the two foci is also a constant.

For this case we can show that the curve satisfies the Two Focus Property with foci at the points (±`c`, 0), where c = √(`a`^{2} − `b`^{2}). Indeed, let (`x`, `y`) be any point satisfying equation (1), and compute the square of its distance from (`c`, 0) and from (−`c`, 0). For the first, the square of the distance is

(2) |

From equation (1), we can write `y`^{2} = `b`^{2 }(1 − `x`^{2}/`a`^{2}) = `b`^{2} − (`b`^{2}/`a`^{2})`x`^{2}. Substitution into Equation (2) then leads to

To simplify this expression, we observe that `c`^{2} + `b`^{2} = `a`^{2}, obtaining

Proceeding further, combine the `x`^{2} terms, and create a common denominator of `a`^{2}. That produces

Finally, substitute `c`^{2} for `a`^{2} − `b`^{2} and recognize a perfect square in the numerator

In this way we derive a simple expression for `d`_{1}. Note though that −`a` ≤ `x ` ≤ `a`, and also 0 < `c` < `a`. Thus `c``x` < ` a`^{2}, so that `d`_{1} = (`a`^{2} − `c``x`)/`a`.

A similar derivation shows that the distance from (`x`, `y`) to (−`c`, 0) is given by `d`_{2} = (`a`^{2} + `c``x`)/`a`. Therefore, the sum of the distances from (`x`, `y`) to the points (−`c`, 0) and (`c`, 0) is 2`a`. This shows that every point on the curve defined by equation (1) satisfies the Two Focus Property with the foci (−`c`, 0) and (`c`, 0) and with the constant `d = `2`a`.

We must also see that no other points satisfy the Two Focus Property with the same foci and constant `d`. To do so, we should begin with a point (`x`, `y`) whose distances from (−`c`, 0) and (`c`, 0) add up to 2`a`, and show that this point satisfies equation (1). This can be accomplished with algebra that is very similar to the steps shown above, and so the details will be omitted.

This case is similar to the previous one, but this time `c` = √(`b`^{2} − `a`^{2}) and the foci are at (0, `c`) and (0, −`c`). Using algebra that is essentially the same as before, you can show that here again the Two Focus Property is obtained.

Overall, then, we have seen that a curve given by equation (1) coincides with the set of points satisfying the Two Focus Property for a specific pair of foci and constant `d`.

Suppose a curve satisfies the Two Focus Property. Impose a coordinate system so that the two foci are on the `x`-axis, with the origin centered between them. Define `c` so that the foci are at (±`c`, 0). This makes the distance between the foci 2`c`, which must be less than `d`. Equivalently, we have `c` < `d`/2.

Next define `a` = `d`/2 (which is greater than `c`) and `b` = √(`a`^{2} − `c`^{2}). From the preceding arguments, we know that the points satisfying equation (1) are exactly the same as those satisfying the Two Focus Property. This shows that the original curve is given by an equation of the same form as equation (1).

We have now seen that the Two Focus Property is just as valid a way to define an ellipse as the method of stretching a circle. Using either approach, we know that an ellipse centered at the origin and with its axes along the coordinate axes satisfies an equation with the same form as equation (1). We also know that the ellipse has foci at points on the longer axis of the ellipse, and that a point is on the ellipse if and only if it satisfies the Two Focus Property for a given constant `d`. The next result we need shows that when the points of an ellipse are acted on by a linear transformation, the resulting curve is still an ellipse.