Find foci, vertices, lengths of major and minor axes and the eccentricity of the ellipse:$$ \dfrac{ \left( x + 6 \right)^2}{ 49 } + \dfrac{ \left( y + 3 \right)^2}{ 64 } = 1 $$
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 + 6 \right)^2}{ 49 } + \dfrac{ \left( y + 3 \right)^2}{ 64 } = 1 $ we conclude that:
$$ h = -6, ~~ k = -3, ~~ a^2 = 49 ~~ b^2 = 64 $$ $$ a = \sqrt{ 49 } = 7 ~~\text{and} ~~ b = \sqrt{ 64 } = 8 $$In this example is $ a < b $, so we will use formulas for this case.
Center is $ (h, k) = ( -6, -3 ) $
Major Axis Length is $2 b = 2 \cdot 8 = 16 $
Minor Axis Length is $2 a = 2 \cdot 7 = 14 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 64 - 49 } = \sqrt{ 15 } $$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ \sqrt{ 15 } }{ 7 } = \frac{\sqrt{ 15 }}{ 7 } $$Area is $ A = a b \pi = 7 \cdot 8 \cdot \pi = 56\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(-6, -3 - \sqrt{ 15 } \right) = \left(-3, -6.873 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(-6, -3 + \sqrt{ 15 }\right) = \left(-3, 0.873 \right) $
First Vertex is $ \left(h, k - b \right) = \left(-6, -3 - 8 \right) = \left(-6, -11 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(-6, -3 + 8 \right) = \left(-6, 5 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(-6 - 7, -3 \right) = \left(-13, -3 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(-6 + 7, -3 \right) = \left(1, -3 \right) $