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 - 3 \right)^2}{ 9 } = 1 $ we conclude that:
$$ h = -3, ~~ k = 3, ~~ a^2 = 36 ~~ b^2 = 9 $$ $$ a = \sqrt{ 36 } = 6 ~~\text{and} ~~ b = \sqrt{ 9 } = 3 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( -3, 3 ) $
Major Axis Length is $2 a = 2 \cdot 6 = 12 $
Minor Axis Length is $2 b = 2 \cdot 3 = 6 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 36 - 9 } = \sqrt{ 27 } = 3 \sqrt{ 3 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 3 \sqrt{ 3 } }{ 6 } = \frac{\sqrt{ 3 }}{ 2 } $$Area is $ A = a b \pi = 6 \cdot 3 \cdot \pi = 18\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(-3 - 3 \sqrt{ 3 }, 3 \right) = \left(-8.1962, 3 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(-3 + 3 \sqrt{ 3 }, 3 \right) = \left(2.1962, 3 \right) $
First Vertex is $ \left(h - a, k \right) = \left(-3 - 6, 3 \right) = \left(-9, 3 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(-3 + 6, 3 \right) = \left(3, 3 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(-3, 3 - 3 \right) = \left(-3, 0 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(-3, 3 + 3 \right) = \left(-3, 6 \right) $