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}{ 100 } + \dfrac{ \left( y - 6 \right)^2}{ 1 } = 1 $ we conclude that:
$$ h = -3, ~~ k = 6, ~~ a^2 = 100 ~~ b^2 = 1 $$ $$ a = \sqrt{ 100 } = 10 ~~\text{and} ~~ b = \sqrt{ 1 } = 1 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( -3, 6 ) $
Major Axis Length is $2 a = 2 \cdot 10 = 20 $
Minor Axis Length is $2 b = 2 \cdot 1 = 2 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 100 - 1 } = \sqrt{ 99 } = 3 \sqrt{ 11 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 3 \sqrt{ 11 } }{ 10 } = \frac{ 3 \sqrt{ 11}}{ 10 } $$Area is $ A = a b \pi = 10 \cdot 1 \cdot \pi = 10\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(-3 - 3 \sqrt{ 11 }, 6 \right) = \left(-12.9499, 6 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(-3 + 3 \sqrt{ 11 }, 6 \right) = \left(6.9499, 6 \right) $
First Vertex is $ \left(h - a, k \right) = \left(-3 - 10, 6 \right) = \left(-13, 6 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(-3 + 10, 6 \right) = \left(7, 6 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(-3, 6 - 1 \right) = \left(-3, 5 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(-3, 6 + 1 \right) = \left(-3, 7 \right) $