Find the eccentricity, center, foci and vertices of the ellipse $\;$ $9x^2 + 4y^2 = 36$ $\;$ and sketch it.
Equation of given ellipse is: $\;$ $9x^2 + 4y^2 = 36$
i.e. $\;$ $\dfrac{x^2}{36/9} + \dfrac{y^2}{36/4} = 1$
i.e. $\;$ $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$ $\;\;\; \cdots \; (1)$
Major axis of the ellipse is along the Y axis.
Comparing equation $(1)$ with the standard equation of the ellipse $\;$ $\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$ $\;$ gives
$b^2 = 4 \implies b = 2$; $\;$ and $\;$ $a^2 = 9 \implies a = 3$
Now, $e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{4}{9}} = \dfrac{\sqrt{5}}{3}$
$\therefore$ $\;$ $a \times e = 3 \times \dfrac{\sqrt{5}}{3} = \sqrt{5}$
Center $= \left(0, 0\right)$
Foci $\left(F_1 \text{ and } F_2\right) = \left(0, \pm ae\right) = \left(0, \pm \sqrt{5}\right)$
Vertices $\left(A_1 \text{ and } A_2\right) = \left(0, \pm a\right) = \left(0, \pm 3\right)$
