Rewrite ellipse equation in standard form
$$ \begin{aligned} \frac{ x^2 }{ \frac{ 3 }{ 2 } } + \frac{ 3 y^2 }{ \frac{ 3 }{ 2 } } & = 1 \\ \frac{ x^2 }{ \frac{ 3 }{ 2 } } + \frac { y^2 }{ \dfrac{ \frac{ 3 }{ 2 } }{ 3 }} & = 1 \\ \frac { x^2 }{ \frac{ 3 }{ 2 } } + \frac { y^2 }{ \frac{ 1 }{ 2 } } & = 1 \end{aligned} $$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 $ x^2 + y^2 = \frac{ 3 }{ 2 } $ we conclude that:
$$ h = 0, ~~ k = 0, ~~ a^2 = \frac{ 3 }{ 2 } ~~ b^2 = \frac{ 1 }{ 2 } $$ $$ a = \sqrt{ \frac{ 3 }{ 2 } } = \frac{\sqrt{ 6 }}{ 2 } ~~\text{and} ~~ b = \sqrt{ \frac{ 1 }{ 2 } } = \frac{\sqrt{ 2 }}{ 2 } $$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 \frac{\sqrt{ 6 }}{ 2 } = \sqrt{ 6 } $
Minor Axis Length is $2 b = 2 \cdot \frac{\sqrt{ 2 }}{ 2 } = \sqrt{ 2 } $
Linear Eccentricity (focal distance) is:
$$ c = \sqrt{a^2 - b^2} = \sqrt{ \frac{ 3 }{ 2 } - \frac{ 1 }{ 2 } } = \sqrt{ 1 } = 1$$Eccentricity is:
$$ e = \dfrac{ c } { a } = \dfrac{ 1 }{ \frac{\sqrt{ 6 }}{ 2 } } = \frac{\sqrt{ 6 }}{ 3 } $$Area is $ A = a b \pi = \frac{\sqrt{ 6 }}{ 2 } \cdot \frac{\sqrt{ 2 }}{ 2 } \cdot \pi = \frac{\sqrt{ 3 }}{ 2 }\pi $
First focus is $ \text{F1} = \left(h - c, k \right) = \left(0 - 1, 0 \right) = \left(-1, 0 \right) $
Second focus is $ \text{F2} =\left(h + c, k \right) = \left(0 + 1, 0 \right) = \left(1, 0 \right) $
First Vertex is $ \left(h - a, k \right) = \left(0 - \frac{\sqrt{ 6 }}{ 2 }, 0 \right) = \left(- \frac{\sqrt{ 6 }}{ 2 }, 0 \right) $
Second Vertex is $ \left(h + a, k \right) = \left(0 + \frac{\sqrt{ 6 }}{ 2 }, 0 \right) = \left(\frac{\sqrt{ 6 }}{ 2 }, 0 \right) $
First Co-vertex is $ \left(h, k - b \right) = \left(0, 0 - \frac{\sqrt{ 2 }}{ 2 } \right) = \left(0, - \frac{\sqrt{ 2 }}{ 2 } \right) $
Second Co-vertex is $ \left(h, k + b \right) = \left(0, 0 + \frac{\sqrt{ 2 }}{ 2 } \right) = \left(0, \frac{\sqrt{ 2 }}{ 2 } \right) $