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 + 5 \right)^2}{ 9 } + \dfrac{ \left( y - 1 \right)^2}{ 4 } = 1 $ we conclude that:
$$ h = -5, ~~ k = 1, ~~ a^2 = 9 ~~ b^2 = 4 $$ $$ a = \sqrt{ 9 } = 3 ~~\text{and} ~~ b = \sqrt{ 4 } = 2 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( -5, 1 ) $
Major Axis Length is $2 a = 2 \cdot 3 = 6 $
Minor Axis Length is $2 b = 2 \cdot 2 = 4 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 9 - 4 } = \sqrt{ 5 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 5 } }{ 3 } = \frac{\sqrt{ 5 }}{ 3 } $$Area is $ A = a b \pi = 3 \cdot 2 \cdot \pi = 6\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(-5 - \sqrt{ 5 }, 1 \right) = \left(-7.2361, 1 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(-5 + \sqrt{ 5 }, 1 \right) = \left(-2.7639, 1 \right) $
First Vertex is $ \left(h - a, k \right) = \left(-5 - 3, 1 \right) = \left(-8, 1 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(-5 + 3, 1 \right) = \left(-2, 1 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(-5, 1 - 2 \right) = \left(-5, -1 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(-5, 1 + 2 \right) = \left(-5, 3 \right) $