Find the coordinates of the foci and the vertices, the eccentricity,
Question:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$

Solution:

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

On comparing this equation with the standard equation of hyperbola i.e., $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we obtain $a=4$ and $b=3 .$

We know that $a^{2}+b^{2}=c^{2}$.

$\therefore c^{2}=4^{2}+3^{2}=25$

$\Rightarrow c=5$

Therefore,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Eccentricity, $e=\frac{c}{a}=\frac{5}{4}$

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