# Find the coordinates of the foci, the vertices, the length of major axis, the minor axis,

Question:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

Solution:

The given equation is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ or $\frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{2}}=1$

Here, the denominator of $\frac{x^{2}}{16}$ is greater than the denominator of $\frac{y^{2}}{9}$.

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we obtain $a=4$ and $b=3$.

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{16-9}=\sqrt{7}$

Therefore,

The coordinates of the foci are $(\pm \sqrt{7}, 0)$.

The coordinates of the vertices are $(\pm 4,0)$

Length of major axis = 2a = 8

Length of minor axis = 2b = 6

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{7}}{4}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}$