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 - 3 \right)^2}{ 36 } + \dfrac{ \left( y + 5 \right)^2}{ 100 } = 1 $ we conclude that:
$$ h = 3, ~~ k = -5, ~~ a^2 = 36 ~~ b^2 = 100 $$ $$ a = \sqrt{ 36 } = 6 ~~\text{and} ~~ b = \sqrt{ 100 } = 10 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( 3, -5 ) $
Major Axis Length is $2 b = 2 \cdot 10 = 20 $
Minor Axis Length is $2 a = 2 \cdot 6 = 12 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 100 - 36 } = \sqrt{ 64 } = 8$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 8 }{ 6 } = \frac{ 4 }{ 3 } $$Area is $ A = a b \pi = 6 \cdot 10 \cdot \pi = 60\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(3, -5 - 8 \right) = \left(-5, -13 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(3, -5 + 8\right) = \left(-5, 3 \right) $
First Vertex is $ \left(h, k - b \right) = \left(3, -5 - 10 \right) = \left(3, -15 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(3, -5 + 10 \right) = \left(3, 5 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(3 - 6, -5 \right) = \left(-3, -5 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(3 + 6, -5 \right) = \left(9, -5 \right) $