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 + 1 \right)^2}{ 4 } + \dfrac{ \left( y - 4 \right)^2}{ 9 } = 1 $ we conclude that:
$$ h = -1, ~~ k = 4, ~~ a^2 = 4 ~~ b^2 = 9 $$ $$ a = \sqrt{ 4 } = 2 ~~\text{and} ~~ b = \sqrt{ 9 } = 3 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( -1, 4 ) $
Major Axis Length is $2 b = 2 \cdot 3 = 6 $
Minor Axis Length is $2 a = 2 \cdot 2 = 4 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 9 - 4 } = \sqrt{ 5 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 5 } }{ 2 } = \frac{\sqrt{ 5 }}{ 2 } $$Area is $ A = a b \pi = 2 \cdot 3 \cdot \pi = 6\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(-1, 4 - \sqrt{ 5 } \right) = \left(4, 1.7639 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(-1, 4 + \sqrt{ 5 }\right) = \left(4, 6.2361 \right) $
First Vertex is $ \left(h, k - b \right) = \left(-1, 4 - 3 \right) = \left(-1, 1 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(-1, 4 + 3 \right) = \left(-1, 7 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(-1 - 2, 4 \right) = \left(-3, 4 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(-1 + 2, 4 \right) = \left(1, 4 \right) $