Given:$\large\frac{x^2}{49}$$+\large\frac{y^2}{36 }$$=1$

Since $49 \gt 36$, he major axis is along the x-axis.

On comparing the equation with $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1 \rightarrow a = \sqrt {49} = 7$ and $b = \sqrt {36} =6$

$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{49-36} = \sqrt{13}$

1) The coordinates of the foci are $(\sqrt{ 13}, 0)$ and $(-\sqrt{ 13}, 0)$

2) The coordinates of the vertex are $(7,0)$ and $(-7,0)$

3) The Length of the major axis, $2a = 2 \times 7= 14$

4) The Length of the minor axis, $2b = 2 \times 3 = 6$

5) The Eccentricity, $e = \large\frac{c}{a} $$ = \large\frac{\sqrt{13}}{7}$

6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$ = \large\frac{2 \times 6^2}{7} $$ = \large\frac{72}{7}$