Ellipses

An ellipse is the set of all points the sum of whose distances from two fixed points (foci) is constant.  

The image above, a demonstration of the definition of an ellipse, shows that if you move the pencil to any point on the ellipse, the total length of the 2 lines connecting the pencil with the 2 points (foci) remains the same.

An interactive program displaying the above demonstration can also be seen on the web page: http://www.mathopenref.com/ellipse.html

Here are some of the parts of an ellipse as portrayed in the picture below the chart:

Standard Form of an Ellipse  

Using an ellipse with center (h, k); major axis length 2a and minor axis 2b; c as the distance between the center and foci:  

The graph below is portraying the standard equation where the ellipse is horizontal because, in reference to the first equation above, a is greater than b. Which ever denominator is larger is a, showing the direction of the major axis. For example, when the quantity (y-k)^2 is over a^2, the major axis runs vertically. When it is over b^2, it runs horizontally.

Likewise, in the graph below, the major axis runs vertically; this is because in correspondence to the second equation, (x-h)^2 is found over b^2 and (y-k)^2 is found over a^2.

In order to find the foci of the standard form of an ellipse, the equation is c^2=a^2-b^2.

Eccentricity, or the "ovalness" of the ellipse, is calculated using e=c/a. The smaller the calculated eccentricity is, the closer the ellipse is to being in a circular form. In comparison, the closer the eccentricity is to 1, the more oval the ellipse is.

Example

Find the center, vertices, co-vertices, foci, and eccentricity of  9x² + 4y²+ 36x - 24y = -36.

     First, change the equation into standard form:

                              9(x² + 4x + 4) + 4(y² – 6y + 9) = 36

                              9(x + 2)² + 4(y - 3)² = 36

                              (x + 2)² + (y - 3)² = 1 

                                  4            9

     Then from the standard equation find the center (h, k) using the (x+ 2)² + (y- 3)², which we know should be the same as (x - h)² + (y - k)², so -h=2 and -k=-3. Therefore, the center is (-2, 3). What we also know from the standard equation is that the graph would be vertical because its denominator is larger than the denominator of x. From this, a²=9, making a = +3,-3. To calculate the vertices we add a to the center according to the direction of the graph. Since we know that this graph is vertical, we add a to the y coordinate in the center for our vertices which turn out to be (-2, 6) and (-2, 0). Then to find the co-vertices we do the same thing, though using b (2 and -2) this time and adding the results to other coordinate, in this case, x. The co-vertices are (0, 3) and (-4, 3). The next thing to calculate is the foci of the ellipse. In order to find these points, we use the equation c^2=a^2-b^2. The equation results with c = +5^(1/2), -5^(1/2). Then we add these numbers to the y coordinate (again because it's vertical). The foci are located at the points (-2, 3+5^(1/2)) and (-2, 3-5^(1/2)). The last part of the problem is to calculate the eccentricity of the ellipse. Using e=c/a, e = .745. The graph of this ellipse is shown below. 

Want to try it for yourself? Ellipse extra practice problems and more guided problems here.

Below is another link that will take you to a video explaining how to determine the equation of an ellipse using its eccentricity, along with a great review of everything learned about ellipses.

Ellipse review