# How do you find the center, vertices, and foci of an ellipse (1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1?

Nov 17, 2015

See explanation

#### Explanation:

Standard equation of an ellipse

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

or

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

where $a > b$

Since the denominator of the $x$ is larger, we use the first equation

$C : \left(h , k\right) \implies \left(- 2 , 5\right)$

$V : \left(h \pm a , k\right) \implies \left(- 2 \pm 4 , 5\right)$

$f : \left(h \pm c , k\right)$

where ${c}^{2} = {a}^{2} - {b}^{2}$

$\implies {c}^{2} = 16 - 9 \implies 7$
$\implies c = \sqrt{7}$

$\implies f : \left(- 2 \pm \sqrt{7} , 5\right)$