# How do you find the equation of an ellipse with foci (+-5,0) and major axis of length 12?

Dec 22, 2016

#### Explanation:

Because the x coordinate of the foci is the coordinate that is changing, we know that the major axis of the ellipse is parallel to the x axis. Therefore, the standard Cartesian form of the equation of the ellipse is:

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

where $\left(h , k\right)$ is the center, "a" is half of the length of the major axis and "b" is half of the length of the minor axis.

We are given that the major axis is length 12, therefore, we substitute 6 for "a" into equation [1]:

${\left(x - h\right)}^{2} / {6}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [2]}$

The foci for this type of ellipse are located at:

$\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) \mathmr{and} \left(h + \sqrt{{a}^{2} - {b}^{2}} , k\right)$

Using the foci, $\left(- 5 , 0\right) \mathmr{and} \left(5 , 0\right)$, and $a = 6$ we can write 3 equations that will help is to find the values of $h , k , \mathmr{and} b$:

$h - \sqrt{{6}^{2} - {b}^{2}} = - 5 \text{ [3]}$
$h + \sqrt{{6}^{2} - {b}^{2}} = 5 \text{ [4]}$
$k = 0 \text{ [5]}$

$2 h = 0$

$h = 0$

Subtract equation [3] from equation [4]:

$2 \sqrt{{6}^{2} - {b}^{2}} = 10$

$\sqrt{{6}^{2} - {b}^{2}} = 5$

${6}^{2} - {b}^{2} = 25$

$- {b}^{2} = 25 - {6}^{2}$

$- {b}^{2} = - 11$

$b = \sqrt{11}$

Substitute these values into equation [2]:

${\left(x - 0\right)}^{2} / {6}^{2} + {\left(y - 0\right)}^{2} / {\left(\sqrt{11}\right)}^{2} = 1 \text{ [6]}$