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}{ 16 } + \dfrac{ \left( y + 1 \right)^2}{ 9 } = 1 $ we conclude that:
$$ h = 4, ~~ k = -1, ~~ a^2 = 16 ~~ b^2 = 9 $$ $$ a = \sqrt{ 16 } = 4 ~~\text{and} ~~ b = \sqrt{ 9 } = 3 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( 4, -1 ) $
Major Axis Length is $2 a = 2 \cdot 4 = 8 $
Minor Axis Length is $2 b = 2 \cdot 3 = 6 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 16 - 9 } = \sqrt{ 7 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 7 } }{ 4 } = \frac{\sqrt{ 7 }}{ 4 } $$Area is $ A = a b \pi = 4 \cdot 3 \cdot \pi = 12\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(4 - \sqrt{ 7 }, -1 \right) = \left(1.3542, -1 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(4 + \sqrt{ 7 }, -1 \right) = \left(6.6458, -1 \right) $
First Vertex is $ \left(h - a, k \right) = \left(4 - 4, -1 \right) = \left(0, -1 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(4 + 4, -1 \right) = \left(8, -1 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(4, -1 - 3 \right) = \left(4, -4 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(4, -1 + 3 \right) = \left(4, 2 \right) $