Ellipse

An ellipse is a conic section. The basic property of an ellipse is that every point on it has the same sum of distances from two given points in the plane. These points are called the foci.

What an ellipse looks like

Let’s take two points – E and F, the foci of the ellipse. Then for every point X on the ellipse,

$$|XE|+|XF|=K,$$

where K is some constant number. This number is therefore the same for all points of the ellipse. If E = F, we get a circle, and |XE|+|XF| equals the diameter of the circle (that is, |XE| must be the radius). Here is a picture of an ellipse:

The same sum of distances from the foci means that |EK|+|FK| must equal |EL|+|FL|, and the same holds for every other point on the ellipse.

Description and properties of an ellipse

• An ellipse has two foci (singular: focus), which we label E and F.
• An ellipse has two vertices on the major axis, A and B, and two vertices on the minor axis, C and D.
• The centre of the ellipse, point S in the picture, lies at the midpoint of the line segment EF, that is, between the foci.
• The line through the vertices on the major axis (and also through the foci) is called the major axis of the ellipse. The line through the vertices on the minor axis is called the minor axis.
• A line segment joining either end of the major axis to the centre is called a semi-major axis. In the picture these are the segments AS and BS.
• A line segment joining either end of the minor axis to the centre is called a semi-minor axis. In the picture these are the segments CS and DS.
• The constant K, the sum of the distances from a point on the ellipse to the two foci, equals the length of AB. This is easy to see if we work out the sum for the point B. For B the sum is |FB|+|EB|. The segment EB covers almost the whole of AB, and the remaining part, AE, is as long as FB. Therefore |FB|+|EB| = |AB|.

Linear eccentricity of an ellipse

Another important constant of an ellipse is the linear eccentricity, written e. It is the distance from each focus to the centre of the ellipse, so e = |ES| = |FS|. (Plain “eccentricity” means the ratio of this length to the semi-major axis.) How can we work out the linear eccentricity? First we find the distances |ED| and |FD|.

We know that the sum K equals |AB|. It is also clear from the picture that the segments ED and FD have the same length, because the line CD is an axis of the ellipse. Each of them is therefore half of |AB|, which is the length of the semi-major axis. If we call the length of the semi-major axis a and the length of the semi-minor axis b, we get this picture:

We can now express the linear eccentricity using Pythagoras’ theorem as

$$e=\sqrt{a^2-b^2}$$

The more an ellipse resembles a circle, that is, the less flattened it is, the smaller its linear eccentricity.

The equation of an ellipse

How do we derive the equation of an ellipse? First we need a nice ellipse, so we choose one whose major and minor axes are parallel to the x-axis and y-axis and whose centre is at the origin of the coordinate system, that is, at (0, 0). Such an ellipse looks like this:

The question is how to describe the point X, marked in blue in the picture, in general. From the definition of an ellipse we know that

$$|EX|+|FX|=|AB|$$

Since the length of AB is twice the length of the semi-major axis a, we can write:

$$|EX|+|FX|=2a$$

Next we need to express the lengths of the segments EX and FX. We start with FX. We add two more segments to the picture so that we get a right-angled triangle.

The point X has coordinates (x, y), so the length of PX equals the y-coordinate of X, that is, |PX| = y. The length of PF is |e − x|. By Pythagoras’ theorem,

$$|FX|=\sqrt{|e-x{|}^2+y^2}=\sqrt{(x-e{)}^2+y^2}$$

In the same way that we expressed |FX|, we can express |EX|. The length of PX is the same, again y, and the length of PE is x + e. We get:

$$|EX|=\sqrt{(x+e{)}^2+y^2}$$

We substitute these new expressions into the previous equation:

$$\sqrt{(x+e{)}^2+y^2}+\sqrt{(x-e{)}^2+y^2}=2a$$

Now there would follow a big pile of algebra, until you finally arrive at this nice form:

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

This equation, however, describes an ellipse centred at the origin. To get a more general equation we have to include the coordinates of the centre. An ellipse with centre (m, n) has the equation:

If the major axis is parallel to the x-axis:

$$\frac{(x-m{)}^2}{a^2}+\frac{(y-n{)}^2}{b^2}=1$$

If the major axis is parallel to the y-axis:

$$\frac{(x-m{)}^2}{b^2}+\frac{(y-n{)}^2}{a^2}=1$$

Example: to convince yourself that the equation really works, let’s substitute the point X from the picture. In the picture a = 5, b = 4 and X(2.31, 3.55) (the coordinates are rounded, so we have to round the result too). Substituting into the left-hand side of the equation, we get:

$$\frac{{2.31}^2}{5^2}+\frac{{3.55}^2}{4^2}=\frac{5.3361}{25}+\frac{12.6025}{16}=0.213444+0.78765625\approx 1$$

So we have obtained the equality 1 = 1.

Calculator

Enter the lengths of the semi-major and semi-minor axes. The area of an ellipse is π · a · b. The perimeter of an ellipse cannot be worked out exactly, so the calculator uses Ramanujan’s approximate formula, which differs from the true value only very slightly.