Ellipse

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

What an ellipse looks like

Given two points - E and F, the foci of the ellipse are e. Then for each point X of the ellipse e it must hold that

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

where K is some constant number. This number is thus the same for all points of the ellipse. In the case of E = F, we get a circle and it holds that |XE|+|XF| is equal to the diameter of the circle (or |XE| must be the radius of the circle). Image of the ellipse:

The equal sum of the distances from the foci then means that the sum of |EK|+|FK| must be the same as the sum of |EL|+|FL|, and the same for all other points that are on the ellipse.

Description and properties of the ellipse

• The ellipse has two foci, let us denote them E and F.
• The ellipse contains two major vertices, A and B, and two minor vertices, C and D.
• The center of the ellipse, vertex S in the figure, lies at the center of the line segment EF, between the foci.
• The line that passes through the major vertices (and also through the foci) is called the major a xis of the ellipse, and the line that passes through the minor vertices is called the minor axis of the ellipse.
• The line that connects any major vertex and the center of the ellipse is called the major semi-axis. In the figure, these are the lines AS and BS.
• The line that connects any minor point and the center of the ellipse is called the minor semi-axis. In the figure, these are the lines CS and DS.
• The constant K, which is equal to the sum of the lengths of the ellipse's connecting points with the foci, is equal to the length of the line segment AB. This is nice to see if we want to calculate the sum for the point B. For the point, the sum is: |FB|+|EB|. The line segment EB covers almost the entire line segment AB, and the remaining part, the line segment AE, is the same length as the line segment FB. Therefore |FB|+|EB| = |AB|.

The eccentricity of the ellipse

Another important constant in an ellipse is eccentricity, denoted e, or eccentricity. Eccentricity is equal to the distance of the foci from the center of the ellipse, i.e. e = |ES| = |FS|. How can we calculate eccentricity? First we find what the distance |ED| and |FD| is equal to .

We know that the sum of K is equal to the length of |AB|. Yet the figure shows that the segments ED and FE will be the same length because the line CD is the axis of the ellipse. Thus the length of the two segments will be equal to half the length of |AB|, which is the length of the major semi-axis of the ellipse. If we denote the length of the major semi-axis as a and the length of the minor semi-axis as b, we get the figure:

The eccentricity can then be expressed using the Pythagorean theorem as

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

The more similar the ellipse is to a circle, i.e. the less flattened it is, the less eccentric it is.

The ellipse equation

How do we derive the equation of an ellipse? The first thing we need is a nice ellipse, so we choose an ellipse whose major and minor axes are parallel to the axes x and y, and the center of the ellipse is at the origin of the coordinate system, i.e., it has coordinates [0, 0]. For example, such an ellipse looks like this:

The question is how to express in general the point X, which is marked in blue in the figure. We know from the definition of an ellipse that the following must hold:

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

Since the length of the line segment AB is equal to twice the length of the main axis a, we can write:

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

Next, we must somehow express the length of the segments EX and FX. We start with the segment FX. Let's draw two more lines in the figure to make a right triangle.

The point X has the coordinates [x, y], so the length of the line segment PX is equal to y-the coordinate of the point X, i.e. |PX| = y. The length of PF is equal to |e − x|. By the Pythagorean theorem, then it must hold:

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

Similarly as we expressed |FX| we express |EX|. The length of PX will be the same, again y and the length of PE will be equal to x + e. We get:

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

We plug these new expressions into the previous equation:

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

Now a big pile of adjustments would follow until you finally get this nice shape:

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

However, this equation assumes an ellipse with the center at the origin. To get a more general equation, we need to include the coordinates of the center of the ellipse in the equation. An ellipse centered at [m, n] has the equation:

If the principal axis is parallel to the axis of x:

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

If the major axis is parallel to the axis y:

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

Example: to believe that the equation actually works, try plugging in the point X from the figure. For the picture, a = 5, b = 4 and X = [2,31; 3,55] apply (the coordinates of the point are rounded, we have to round the result in the end as well). By substituting in the left part of the equation, we get:

$$\frac{2{,}31^2}{5^2}+\frac{3{,}55^2}{4^2}=\frac{5{,}3361}{25}+\frac{12{,}6025}{16}=0{,}213444+0{,}78765625\approx1$$

We have thus obtained the equality 1 = 1.