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 - 0 \right)^2}{ 30 } + \dfrac{ \left( y - 0 \right)^2}{ 5 } = 1 $ we conclude that:
$$ h = 0, ~~ k = 0, ~~ a^2 = 30 ~~ b^2 = 5 $$ $$ a = \sqrt{ 30 } ~~\text{and} ~~ b = \sqrt{ 5 } $$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 a = 2 \cdot \sqrt{ 30 } = 2 \sqrt{ 30 } $
Minor Axis Length is $2 b = 2 \cdot \sqrt{ 5 } = 2 \sqrt{ 5 } $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 30 - 5 } = \sqrt{ 25 } = 5$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 5 }{ \sqrt{ 30 } } = \frac{\sqrt{ 30 }}{ 6 } $$Area is $ A = a b \pi = \sqrt{ 30 } \cdot \sqrt{ 5 } \cdot \pi = 5 \sqrt{ 6 }\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(0 - 5, 0 \right) = \left(-5, 0 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(0 + 5, 0 \right) = \left(5, 0 \right) $
First Vertex is $ \left(h - a, k \right) = \left(0 - \sqrt{ 30 }, 0 \right) = \left(-\sqrt{ 30 }, 0 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(0 + \sqrt{ 30 }, 0 \right) = \left(\sqrt{ 30 }, 0 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(0, 0 - \sqrt{ 5 } \right) = \left(0, -\sqrt{ 5 } \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(0, 0 + \sqrt{ 5 } \right) = \left(0, \sqrt{ 5 } \right) $