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