Find foci, vertices, lengths of major and minor axes and the eccentricity of the ellipse:$$ \dfrac{ \left( x - 1 \right)^2}{ 4 } + \dfrac{ \left( y + 2 \right)^2}{ 16 } = 1 $$
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 - 1 \right)^2}{ 4 } + \dfrac{ \left( y + 2 \right)^2}{ 16 } = 1 $ we conclude that:
$$ h = 1, ~~ k = -2, ~~ a^2 = 4 ~~ b^2 = 16 $$ $$ a = \sqrt{ 4 } = 2 ~~\text{and} ~~ b = \sqrt{ 16 } = 4 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( 1, -2 ) $
Major Axis Length is $2 b = 2 \cdot 4 = 8 $
Minor Axis Length is $2 a = 2 \cdot 2 = 4 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 16 - 4 } = \sqrt{ 12 } = 2 \sqrt{ 3 }$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 2 \sqrt{ 3 } }{ 2 } = \sqrt{ 3 } $$Area is $ A = a b \pi = 2 \cdot 4 \cdot \pi = 8\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(1, -2 - 2 \sqrt{ 3 } \right) = \left(-2, -5.4641 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(1, -2 + 2 \sqrt{ 3 }\right) = \left(-2, 1.4641 \right) $
First Vertex is $ \left(h, k - b \right) = \left(1, -2 - 4 \right) = \left(1, -6 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(1, -2 + 4 \right) = \left(1, 2 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(1 - 2, -2 \right) = \left(-1, -2 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(1 + 2, -2 \right) = \left(3, -2 \right) $