Question

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $16x^ 2 - 9y^2 = 576$

Answer 1

The given equation is $16x ^2 - 9y ^2 = 576.$
It can be written as $16x^ 2 - 9y^ 2 = 576$
or $\frac{x^2}{36} –\frac{ y^2}{64} = 1$

or $\frac{x^2}{6^2} – \frac{y^2}{8^2} = 1.......(1)$

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

$∴ c^2 = 36 + 64$
$c = \sqrt{100}$
$c = 10$

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 $=\frac{ 2b^2}{a} = \frac{(2 \times 82)}{6} = \frac{(2\times64)}{6} = \frac{64}{3}$