The standard equation of an ellipse is $ \dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1 $. Comparing to the equation of our ellipse $ \dfrac{ \left( x + 8 \right)^2}{ 25 } + \dfrac{ \left( y + 9 \right)^2}{ 16 } = 1 $ we conclude that:
$$ h = -8, ~~ k = -9, ~~ a^2 = 25 ~~ b^2 = 16 $$ $$ a = \sqrt{ 25 } = 5 ~~\text{and} ~~ b = \sqrt{ 16 } = 4 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( -8, -9 ) $
Major Axis Length is $2 a = 2 \cdot 5 = 10 $
Minor Axis Length is $2 b = 2 \cdot 4 = 8 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 25 - 16 } = \sqrt{ 9 } = 3$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 3 }{ 5 } = \frac{ 3 }{ 5 } $$Area is $ A = a b \pi = 5 \cdot 4 \cdot \pi = 20\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(-8 - 3, -9 \right) = \left(-11, -9 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(-8 + 3, -9 \right) = \left(-5, -9 \right) $
First Vertex is $ \left(h - a, k \right) = \left(-8 - 5, -9 \right) = \left(-13, -9 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(-8 + 5, -9 \right) = \left(-3, -9 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(-8, -9 - 4 \right) = \left(-8, -13 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(-8, -9 + 4 \right) = \left(-8, -5 \right) $