# How do you write an equation of an ellipse in standard form given foci at (- 6, 5) and (- 6, - 7) and whose major axis has length 26?

Nov 2, 2017

The equation is ${\left(y + 1\right)}^{2} / 169 + {\left(x + 6\right)}^{2} / 133 = 1$

#### Explanation:

The major axis is vertical

The foci are

$F = \left(- 6 , 5\right) = \left({x}_{0} , {y}_{0} + c\right)$

$F ' = \left(- 6 , - 7\right) = \left({x}_{0} , {y}_{0} - c\right)$

Therefore,

${x}_{0} = - 6$

${y}_{0} = - 1$

The center of the ellipse is $C = \left({x}_{0} , {y}_{0}\right) = \left(- 6 , - 1\right)$

Also,

$2 a = 26$, $\implies$, $a = 13$

$F F ' = 2 c = 12$, $\implies$, $c = 6$

Therefore,

${b}^{2} = {a}^{2} - {c}^{2} = {13}^{2} - {6}^{2} = 169 - 36 = 133$

$b = \sqrt{133}$

The equation of the ellipse is

${\left(y - {y}_{0}\right)}^{2} / {a}^{2} + {\left(x - {x}_{0}\right)}^{2} / {b}^{2} = 1$

${\left(y + 1\right)}^{2} / 169 + {\left(x + 6\right)}^{2} / 133 = 1$

graph{((y+1)^2/169+(x+6)^2/133-1)(y-1000(x+6))=0 [-36.9, 14.42, -12.38, 13.3]}