Find foci, vertices, lengths of major and minor axes and the eccentricity of the ellipse:$$ \dfrac{ x^2}{ 144 } + \dfrac{ y^2}{ 169 } = 1 $$
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}{ 144 } + \dfrac{ \left( y - 0 \right)^2}{ 169 } = 1 $ we conclude that:
$$ h = 0, ~~ k = 0, ~~ a^2 = 144 ~~ b^2 = 169 $$ $$ a = \sqrt{ 144 } = 12 ~~\text{and} ~~ b = \sqrt{ 169 } = 13 $$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 b = 2 \cdot 13 = 26 $
Minor Axis Length is $2 a = 2 \cdot 12 = 24 $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{b^2 - a^2} = \sqrt{ 169 - 144 } = \sqrt{ 25 } = 5$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 5 }{ 12 } = \frac{ 5 }{ 12 } $$Area is $ A = a b \pi = 12 \cdot 13 \cdot \pi = 156\pi $
First focus is $ \text{F1} = \left(h, k-c \right) = \left(0, 0 - 5 \right) = \left(0, -5 \right) $
Second focus is $ \text{F2} = \left(h, k + c \right) = \left(0, 0 + 5\right) = \left(0, 5 \right) $
First Vertex is $ \left(h, k - b \right) = \left(0, 0 - 13 \right) = \left(0, -13 \right) $
Second Vertex is $ \left(h, k + b \right) = \left(0, 0 + 13 \right) = \left(0, 13 \right) $
First Co-vertex is $ \left(h - a, k \right) = \left(0 - 12, 0 \right) = \left(-12, 0 \right) $
Second Co-vertex is $ \left(h + a, k \right) = \left(0 + 12, 0 \right) = \left(12, 0 \right) $