**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:

__Part__ |
__Definition__ |

**Major Axis** |
longest length; goes through center and foci |

**Vertices** |
Endpoints of the major axis |

**Minor Axis** |
Shorter length; perpendicular to the major axis |

**Co-Vertices** |
End points of the minor axis |

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 ** **9x**^{2} + 4y^{2}+ 36x - 24y = -36.

First, change the equation into standard form:

**9(x**^{2} + 4x + 4) + 4(**y**^{2} – 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, **a**^{2}=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

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