# 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 $16 x^{2}-9 y^{2}=576$

Solution:

The given equation is $16 x^{2}-9 y^{2}=576$

It can be written as

$16 x^{2}-9 y^{2}=576$

$\Rightarrow \frac{x^{2}}{36}-\frac{y^{2}}{64}=1$

$\Rightarrow \frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$ $\ldots(1)$

On comparing equation (1) with the standard equation of hyperbola i.e., $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we obtain $a=6$ and $b=8$.

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

$\therefore c^{2}=36+64=100$

$\Rightarrow c=10$

Therefore,

The coordinates of the foci are (±10, 0).

The coordinates of the vertices are (±6, 0).

Eccentricity, $e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 64}{6}=\frac{64}{3}$