Find foci, vertices, lengths of major and minor axes and the eccentricity of the ellipse:$$ \dfrac{ x^2}{ 36 } + \dfrac{ y^2}{ 16 } = 1 $$
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}{ 36 } + \dfrac{ \left( y - 0 \right)^2}{ 16 } = 1 $ we conclude that:
$$ h = 0, ~~ k = 0, ~~ a^2 = 36 ~~ b^2 = 16 $$ $$ a = \sqrt{ 36 } = 6 ~~\text{and} ~~ b = \sqrt{ 16 } = 4 $$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 6 = 12 $
Minor Axis Length is $2 b = 2 \cdot 4 = 8 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ 36 - 16 } = \sqrt{ 20 } = 2 \sqrt{ 5 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 2 \sqrt{ 5 } }{ 6 } = \frac{\sqrt{ 5 }}{ 3 } $$Area is $ A = a b \pi = 6 \cdot 4 \cdot \pi = 24\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(0 - 2 \sqrt{ 5 }, 0 \right) = \left(-2 \sqrt{ 5 }, 0 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(0 + 2 \sqrt{ 5 }, 0 \right) = \left(2 \sqrt{ 5 }, 0 \right) $
First Vertex is $ \left(h - a, k \right) = \left(0 - 6, 0 \right) = \left(-6, 0 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(0 + 6, 0 \right) = \left(6, 0 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(0, 0 - 4 \right) = \left(0, -4 \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(0, 0 + 4 \right) = \left(0, 4 \right) $