# How do you find an equation for the ellipse with vertices at (-6,4) and (10,4); focus at (8,4)?

Nov 16, 2016

Please see the explanation.

#### Explanation:

The equation of an ellipse:

(x - h)^2/a^2 + (y - k)^2/b^2 = 1; a > b

Has vertices at $\left(h \pm a , k\right)$
Has foci at $\left(h \pm \sqrt{{a}^{2} - {b}^{2}} , k\right)$

Use the vertices to write 3 equations:

$k = 4 \text{ [1]}$
$h - a = - 6 \text{ [2]}$
$h + a = 10 \text{ [3]}$

Use equations [2] and [3] to solve for h and a:

$2 h = 4$
$h = 2$
$a = 8$

Use the focus to write another equation:

$8 = h + \sqrt{{a}^{2} - {b}^{2}}$

Substitute values for h and a:

$8 = 2 + \sqrt{{8}^{2} - {b}^{2}}$

$6 = \sqrt{64 - {b}^{2}}$

$36 = 64 - {b}^{2}$

${b}^{2} = 64 - 36$

${b}^{2} = 28$

$b = \sqrt{28}$

Substitute the values into the standard form:

${\left(x - 2\right)}^{2} / {8}^{2} + {\left(y - 4\right)}^{2} / {\left(\sqrt{28}\right)}^{2} = 1$