Rewrite ellipse equation in standard form
$$ \begin{aligned} \frac{ 9 x^2 }{ 81 } + \frac{ y^2 }{ 81 } & = 1 \\ \frac { x^2 }{ \dfrac{ 81 }{ 9 }} + \frac{ y^2 }{ 81 } & = 1 \\ \frac { x^2 }{ 9 } + \frac { y^2 }{ 81 } & = 1 \end{aligned} $$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 $ x^2 + y^2 = 81 $ we conclude that:
$$ h = 0, ~~ k = 0, ~~ a^2 = 9 ~~ b^2 = 81 $$ $$ a = \sqrt{ 9 } = 3 ~~\text{and} ~~ b = \sqrt{ 81 } = 9 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( 0, 0 ) $
Major Axis Length is $2 b = 2 \cdot 9 = 18 $
Minor Axis Length is $2 a = 2 \cdot 3 = 6 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 81 - 9 } = \sqrt{ 72 } = 6 \sqrt{ 2 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 6 \sqrt{ 2 } }{ 3 } = 2 \sqrt{ 2 } $$Area is $ A = a b \pi = 3 \cdot 9 \cdot \pi = 27\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(0, 0 - 6 \sqrt{ 2 } \right) = \left(0, -6 \sqrt{ 2 } \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(0, 0 + 6 \sqrt{ 2 }\right) = \left(0, 6 \sqrt{ 2 } \right) $
First Vertex is $ \left(h, k - b \right) = \left(0, 0 - 9 \right) = \left(0, -9 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(0, 0 + 9 \right) = \left(0, 9 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(0 - 3, 0 \right) = \left(-3, 0 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(0 + 3, 0 \right) = \left(3, 0 \right) $