(N/A) The given equation of the ellipse is $9x^{2} + 4y^{2} = 36$. Dividing both sides by $36$,we get the standard form:
$\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
Since the denominator of $\frac{y^{2}}{9}$ is larger than the denominator of $\frac{x^{2}}{4}$,the major axis is along the $y$-axis.
Comparing this with the standard equation $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,we have $b^{2} = 4$ and $a^{2} = 9$,so $b = 2$ and $a = 3$.
The value of $c$ is given by $c = \sqrt{a^{2} - b^{2}} = \sqrt{9 - 4} = \sqrt{5}$.
The eccentricity $e$ is $e = \frac{c}{a} = \frac{\sqrt{5}}{3}$.
The foci are $(0, \pm c) = (0, \pm \sqrt{5})$.
The vertices are $(0, \pm a) = (0, \pm 3)$.
The length of the major axis is $2a = 2(3) = 6$ units.
The length of the minor axis is $2b = 2(2) = 4$ units.