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 + 6 \right)^2}{ 100 } + \dfrac{ \left( y + 2 \right)^2}{ 15 } = 1 $ we conclude that:
$$ h = -6, ~~ k = -2, ~~ a^2 = 100 ~~ b^2 = 15 $$ $$ a = \sqrt{ 100 } = 10 ~~\text{and} ~~ b = \sqrt{ 15 } $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( -6, -2 ) $
Major Axis Length is $2 a = 2 \cdot 10 = 20 $
Minor Axis Length is $2 b = 2 \cdot \sqrt{ 15 } = 2 \sqrt{ 15 } $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 100 - 15 } = \sqrt{ 85 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 85 } }{ 10 } = \frac{\sqrt{ 85 }}{ 10 } $$Area is $ A = a b \pi = 10 \cdot \sqrt{ 15 } \cdot \pi = 10 \sqrt{ 15 }\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(-6 - \sqrt{ 85 }, -2 \right) = \left(-15.2195, -2 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(-6 + \sqrt{ 85 }, -2 \right) = \left(3.2195, -2 \right) $
First Vertex is $ \left(h - a, k \right) = \left(-6 - 10, -2 \right) = \left(-16, -2 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(-6 + 10, -2 \right) = \left(4, -2 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(-6, -2 - \sqrt{ 15 } \right) = \left(-6, -5.873 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(-6, -2 + \sqrt{ 15 } \right) = \left(-6, 1.873 \right) $