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 $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$.

The given equation is $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$,which can be written as $\frac{x^{2}}{7^{2}}+\frac{y^{2}}{6^{2}}=1$.
Since the denominator of $\frac{x^{2}}{49}$ is greater than the denominator of $\frac{y^{2}}{36}$,the major axis is along the $x$-axis.
Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a=7$ and $b=6$.
We calculate $c = \sqrt{a^{2}-b^{2}} = \sqrt{49-36} = \sqrt{13}$.
$1$. The coordinates of the foci are $(\pm \sqrt{13}, 0)$.
$2$. The coordinates of the vertices are $(\pm 7, 0)$.
$3$. The length of the major axis is $2a = 2 \times 7 = 14$.
$4$. The length of the minor axis is $2b = 2 \times 6 = 12$.
$5$. The eccentricity $e = \frac{c}{a} = \frac{\sqrt{13}}{7}$.
$6$. The length of the latus rectum is $\frac{2b^{2}}{a} = \frac{2 \times 36}{7} = \frac{72}{7}$.