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 - 9 \right)^2}{ 81 } + \dfrac{ \left( y - 4 \right)^2}{ 49 } = 1 $ we conclude that:
$$ h = 9, ~~ k = 4, ~~ a^2 = 81 ~~ b^2 = 49 $$ $$ a = \sqrt{ 81 } = 9 ~~\text{and} ~~ b = \sqrt{ 49 } = 7 $$In this example is $ a > b $, so we will use formulas for this case.
Center is $ (h, k) = ( 9, 4 ) $
Major Axis Length is $2 a = 2 \cdot 9 = 18 $
Minor Axis Length is $2 b = 2 \cdot 7 = 14 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 81 - 49 } = \sqrt{ 32 } = 4 \sqrt{ 2 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 4 \sqrt{ 2 } }{ 9 } = \frac{ 4 \sqrt{ 2}}{ 9 } $$Area is $ A = a b \pi = 9 \cdot 7 \cdot \pi = 63\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(9 - 4 \sqrt{ 2 }, 4 \right) = \left(3.3431, 4 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(9 + 4 \sqrt{ 2 }, 4 \right) = \left(14.6569, 4 \right) $
First Vertex is $ \left(h - a, k \right) = \left(9 - 9, 4 \right) = \left(0, 4 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(9 + 9, 4 \right) = \left(18, 4 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(9, 4 - 7 \right) = \left(9, -3 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(9, 4 + 7 \right) = \left(9, 11 \right) $