# How do you find the foci, center, vertex for (1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1?

$F o c i : {F}_{1} \left(- 2 + \sqrt{7} , 5\right) , {F}_{2} \left(- 2 - \sqrt{7} , 5\right)$ and $C e n t e r : \left(- 2 , 5\right)$
$V e r t e x : {V}_{1} \left(2 , 5\right) , {V}_{2} \left(- 6 , 5\right)$

#### Explanation:

From the given equation of ellipse:

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

From ${\left(x + 2\right)}^{2} / 16 + {\left(y - 5\right)}^{2} / 9 = 1$

it follows that

${\left(x - - 2\right)}^{2} / {4}^{2} + {\left(y - 5\right)}^{2} / {3}^{2} = 1$

And the $C e n t e r \left(h , k\right) = \left(- 2 , 5\right)$

Compute distance from center to focus $c$

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

Compute $F o c i$ ${F}_{1} , {F}_{2}$

${F}_{1} \left(- 2 + \sqrt{7} , 5\right)$

${F}_{2} \left(- 2 - \sqrt{7} , 5\right)$

Compute $V e r t i c e s$ ${V}_{1} , {V}_{2}$

${V}_{1} \left(- 2 + a , 5\right) = {V}_{1} \left(- 2 + 4 , 5\right) = {V}_{1} \left(2 , 5\right)$

${V}_{2} \left(- 2 - a , 5\right) = {V}_{2} \left(- 2 - 4 , 5\right) = {V}_{2} \left(- 6 , 5\right)$

Have a nice day !! from the Philippines!