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?

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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?

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Answer 1 · 2018-01-20T10:35:56

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

Explanation:

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

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

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

$\frac{1}{2} = \frac{c}{5} \to c = \frac{5}{2}$ $1/2=c/5 rarr c = 5/2$ .

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

$b^{2} = 5^{2} - {(\frac{5}{2})}^{2} \to b^{2} = 25 - \frac{25}{4} \to b^{2} = \frac{75}{4}$ $b^2=5^2-(5/2)^2 rarr b^2=25-25/4 rarr b^2=75/4$ .

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

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

So our equation is:

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

or

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

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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?

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