# How do you find the standard form of the equation of the ellipse given the properties center (5,2), vertex (0,2) and eccentricity 1/2?

Jan 20, 2018

${\left(x - 5\right)}^{2} / 25 + \frac{4 {\left(y - 2\right)}^{2}}{75} = 1$

#### Explanation:

Since the center is $\left(5 , 2\right)$ and the vertex is $\left(0 , 2\right)$ we know two things:

$\rightarrow$ $a = 5$, because it's the distance from the center to the vertex.
$\rightarrow$ the ellipse has a horizontal major axis because the center and vertex are on the same horizontal line.

We can calculate $c$, the distance from the center to the focus, because we know $a$ and the eccentricity, $e$, is $e = \frac{c}{a}$:

$\frac{1}{2} = \frac{c}{5} \rightarrow c = \frac{5}{2}$.

In an ellipse we know that ${c}^{2} = {a}^{2} - {b}^{2}$ and $a = 5$ while $c = \frac{5}{2}$, so we can find $b$:

${b}^{2} = {5}^{2} - {\left(\frac{5}{2}\right)}^{2} \rightarrow {b}^{2} = 25 - \frac{25}{4} \rightarrow {b}^{2} = \frac{75}{4}$.

Since the ellipse has a horizontal major axis the standard form of the equation looks like:

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

So our equation is:

${\left(x - 5\right)}^{2} / 25 + {\left(y - 2\right)}^{2} / \left(\frac{75}{4}\right) = 1$

or

${\left(x - 5\right)}^{2} / 25 + \frac{4 {\left(y - 2\right)}^{2}}{75} = 1$