The picture illustrates the definition. The point P is a typical point on the ellipse. The sum of the distances from P to each of the foci is a constant. If we denote this constant by 2a, a > 0, then |PF1| + |PF2| = 2a for any point P on the ellipse.

Other than the foci there are other special points associated with an ellipse which we have pointed out in the diagram. There are also two special line segments associated with an ellipse. The line segment through the foci whose endpoints lie on the ellipse is called the major axis. In the diagram this is AB. The point midway between the foci and lying on the major axis is called the center of the ellipse. This is the point labeled O in the diagram. The points A and B are called the vertices. The line segment through the center and perpendicular to the major axis with endpoints on the ellipse is called the minor axis and is the line segment CD in the diagram.

If we position an ellipse in the plane with its center at the origin and its foci along the x axis we can obtain a nice equation for an ellipse. Let the positive constant be 2a and let c be a positive number so that the foci are located at (-c,0) and (c,0). We have

The numbers a and b do have geometric meaning. Referring to the diagram, we know that at the points A and B, y=0. Then, substituting y=0 in the equation of the ellipse gives x²/a² = 1 which simplifies to x²=a². This equation has solutions x=a or -a. Since a > 0, then A has coordinates (-a,0) and B has coordinates (a,0). Thus the length of AB, which is the length of the major axis is 2a. Similarly, C and D have coordinates (0,b) and (0,-b) respectively. The length of CD, i.e. the length of the minor axis, is 2b. Note that since b² = a²-c² then b < a and so 2b < 2a. In other words, the major axis is always greater in length than the minor axis (hence the choice of the words major and minor).

In a similar manner, we could have placed the foci on the y axis and the center at the origin In this case, the equation of the ellipse comes out to be

The equations we have just established are known as standard equations of an ellipse in standard position. Standard position always implies the center is at the origin and the foci are on one of the axes.

Ellipses in standard position

In this demonstration you can alter the location of the foci and the value of a by moving the sliders. Recall that 2a is the sum of the distances of a point on the ellipse to each foci. Initially the foci are on the x axis but you can also elect to place them on the y axis.

Now suppose we are just given the equation

Example

Find the lengths of the major and minor axes of the ellipse x2/16+y2/25 = 1. and determine the location of the foci. Since 16 < 25 then the major axis is along the y axis. The length of the major axis is 10 (twice the square root of 25) and the length of the minor axis is 8 (twice the square root of 16). The foci are located at (0,c) and (0,-c) where c is a positive number satisfying b2 = a2-c2, or c2 = a2-b2. Then c2 = 25-16 = 9 and so c = 3. The foci are located at (0,3) and (0,-3).

Equation of an ellipse

In this exercise set, you are required to find the equation of an ellipse from given information. You will be given a focus and the length of the major or minor axis or you will be given the lengths of the axes and axis that contains the foci. After you have entered your answer, press "CHECK".

If we translate an ellipse in standard position so that its center is moved to (h, k) then the equation of the ellipse is given as follows:

This form of the equation of an ellipse is called the standard equation. However, if we take a standard equation (x-h)²/u²+(y-k)²/v² = 1 and expand it we get an equation of the form Ax² +By² +Cx +Dy +E=0, where A, B, C, D and E are constants. The question we have is if we are given such an equation can we recognize it as the equation of an ellipse? The answer to this question can be determined by by completing the square. This is similar to the process we used in recognizing a translated parabola. However, unlike that case, we need to apply the technique of completing the square on both the x and y terms.

Example

Determine if 4x2 +y2 -16x+2y+13=0 is the equation of an ellipse. If so, find the coordinates of the center and the foci. Also determine the lengths of the major and minor axes. Our first task is to group together the x terms and group together the y terms. We have 4x2 +y2 -16x+2y+13=0 (4x2 -16x) +(y2 +2y)+13=0 4(x2 -4x) +(y2 +2y)+13=0 Now complete the square on the x terms and the y terms. 4(x2 -4x+4-4) +(y2 +2y+1-1)+13=0 4((x-2)2 -4) +((y+1)2 -1)+13=0 Next we remove the outermost parentheses in the grouped terms. 4(x-2)2 -16 +(y+1)2 -1+13=0 4(x-2)2 +(y+1)2 -4=0 4(x-2)2 +(y+1)2 =4 (x-2)2 +(y+1)2/4 =1 This we should recognize as an ellipse with center (2,-1). Since 4 > 1 and 4 is dividing the "y" term then the major axis is parallel to the y axis. Now, c2 = 4 -1 = 3, so that c = 31/2. The foci of the ellipse in standard position are at (0,c) and (0,-c) so the foci of the translated ellipse are at (2,c-1) and (2,-c-1). That is, at (2,31/2-1) and (2,-31/2-1). Numerically these are approximately (2,0.732) and (2,-2.732). The length of the major axis is 4 (twice the square root of 4) and the length of the minor axis is 2 (twice the square root of 1).

Finding the Center and Foci of an Ellipse

In this exercise you are given the equation of an ellipse in expanded form. On paper you should determine the location of the center and the foci. We suggest you first rewrite the equation in standard form. To accomplish this, you will probably need to complete the square on the x terms and the y terms. Once you have your answer try to sketch the graph of the ellipse. Finally, press the "Graph" button to see the answers.