Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$.

The given equation is $36 x^{2}+4 y^{2}=144$.
Dividing both sides by $144$,we get:
$\frac{36 x^{2}}{144} + \frac{4 y^{2}}{144} = 1$
$\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$
$\frac{x^{2}}{2^{2}} + \frac{y^{2}}{6^{2}} = 1$ ........ $(1)$
Here,the denominator of $\frac{y^{2}}{6^{2}}$ is greater than the denominator of $\frac{x^{2}}{2^{2}}$.
Therefore,the major axis is along the $y$-axis,while the minor axis is along the $x$-axis.
Comparing equation $(1)$ with $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$,we obtain $b = 2$ and $a = 6$.
$\therefore c = \sqrt{a^{2} - b^{2}} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$.
Therefore:
The coordinates of the foci are $(0, \pm 4\sqrt{2})$.
The coordinates of the vertices are $(0, \pm 6)$.
Length of the major axis $= 2a = 12$.
Length of the minor axis $= 2b = 4$.
Eccentricity,$e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$.
Length of the latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 4}{6} = \frac{4}{3}$.