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 + 7 \right)^2}{ 1 } + \dfrac{ \left( y - 3 \right)^2}{ 4 } = 1 $ we conclude that:
$$ h = -7, ~~ k = 3, ~~ a^2 = 1 ~~ b^2 = 4 $$ $$ a = \sqrt{ 1 } = 1 ~~\text{and} ~~ b = \sqrt{ 4 } = 2 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( -7, 3 ) $
Major Axis Length is $2 b = 2 \cdot 2 = 4 $
Minor Axis Length is $2 a = 2 \cdot 1 = 2 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 4 - 1 } = \sqrt{ 3 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 3 } }{ 1 } = \sqrt{ 3 } $$Area is $ A = a b \pi = 1 \cdot 2 \cdot \pi = 2\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(-7, 3 - \sqrt{ 3 } \right) = \left(3, 1.2679 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(-7, 3 + \sqrt{ 3 }\right) = \left(3, 4.7321 \right) $
First Vertex is $ \left(h, k - b \right) = \left(-7, 3 - 2 \right) = \left(-7, 1 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(-7, 3 + 2 \right) = \left(-7, 5 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(-7 - 1, 3 \right) = \left(-8, 3 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(-7 + 1, 3 \right) = \left(-6, 3 \right) $