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}{ 14400 } + \dfrac{ \left( y - 0 \right)^2}{ 4096 } = 1 $ we conclude that:
$$ h = 0, ~~ k = 0, ~~ a^2 = 14400 ~~ b^2 = 4096 $$ $$ a = \sqrt{ 14400 } = 120 ~~\text{and} ~~ b = \sqrt{ 4096 } = 64 $$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 120 = 240 $
Minor Axis Length is $2 b = 2 \cdot 64 = 128 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 14400 - 4096 } = \sqrt{ 10304 } = 8 \sqrt{ 161 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 8 \sqrt{ 161 } }{ 120 } = \frac{\sqrt{ 161 }}{ 15 } $$Area is $ A = a b \pi = 120 \cdot 64 \cdot \pi = 7680\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(0 - 8 \sqrt{ 161 }, 0 \right) = \left(-8 \sqrt{ 161 }, 0 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(0 + 8 \sqrt{ 161 }, 0 \right) = \left(8 \sqrt{ 161 }, 0 \right) $
First Vertex is $ \left(h - a, k \right) = \left(0 - 120, 0 \right) = \left(-120, 0 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(0 + 120, 0 \right) = \left(120, 0 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(0, 0 - 64 \right) = \left(0, -64 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(0, 0 + 64 \right) = \left(0, 64 \right) $