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 - 18 \right)^2}{ \frac{ 393 }{ 100 } } + \dfrac{ \left( y + 12 \right)^2}{ 400 } = 1 $ we conclude that:
$$ h = 18, ~~ k = -12, ~~ a^2 = \frac{ 393 }{ 100 } ~~ b^2 = 400 $$ $$ a = \sqrt{ \frac{ 393 }{ 100 } } = \frac{\sqrt{ 393 }}{ 10 } ~~\text{and} ~~ b = \sqrt{ 400 } = 20 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( 18, -12 ) $
Major Axis Length is $2 b = 2 \cdot 20 = 40 $
Minor Axis Length is $2 a = 2 \cdot \frac{\sqrt{ 393 }}{ 10 } = \frac{\sqrt{ 393 }}{ 5 } $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 400 - \frac{ 393 }{ 100 } } = \sqrt{ \frac{ 39607 }{ 100 } } = \frac{\sqrt{ 39607 }}{ 10 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \frac{\sqrt{ 39607 }}{ 10 } }{ \frac{\sqrt{ 393 }}{ 10 } } = \frac{\sqrt{ 15565551 }}{ 393 } $$Area is $ A = a b \pi = \frac{\sqrt{ 393 }}{ 10 } \cdot 20 \cdot \pi = 2 \sqrt{ 393 }\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(18, -12 - \frac{\sqrt{ 39607 }}{ 10 } \right) = \left(-12, -31.9015 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(18, -12 + \frac{\sqrt{ 39607 }}{ 10 }\right) = \left(-12, 7.9015 \right) $
First Vertex is $ \left(h, k - b \right) = \left(18, -12 - 20 \right) = \left(18, -32 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(18, -12 + 20 \right) = \left(18, 8 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(18 - \frac{\sqrt{ 393 }}{ 10 }, -12 \right) = \left(16.0176, -12 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(18 + \frac{\sqrt{ 393 }}{ 10 }, -12 \right) = \left(19.9824, -12 \right) $