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 + 4 \right)^2}{ 25 } + \dfrac{ \left( y - 4 \right)^2}{ 4 } = 1 $ we conclude that:
$$ h = -4, ~~ k = 4, ~~ a^2 = 25 ~~ b^2 = 4 $$ $$ a = \sqrt{ 25 } = 5 ~~\text{and} ~~ b = \sqrt{ 4 } = 2 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( -4, 4 ) $
Major Axis Length is $2 a = 2 \cdot 5 = 10 $
Minor Axis Length is $2 b = 2 \cdot 2 = 4 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 25 - 4 } = \sqrt{ 21 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 21 } }{ 5 } = \frac{\sqrt{ 21 }}{ 5 } $$Area is $ A = a b \pi = 5 \cdot 2 \cdot \pi = 10\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(-4 - \sqrt{ 21 }, 4 \right) = \left(-8.5826, 4 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(-4 + \sqrt{ 21 }, 4 \right) = \left(0.5826, 4 \right) $
First Vertex is $ \left(h - a, k \right) = \left(-4 - 5, 4 \right) = \left(-9, 4 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(-4 + 5, 4 \right) = \left(1, 4 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(-4, 4 - 2 \right) = \left(-4, 2 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(-4, 4 + 2 \right) = \left(-4, 6 \right) $