Question

Find the coordinates the foci, the vertices, the eccentricity and the length of the latus rectum of the hyperbola,

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

Answer 1

The given equation of hyperbola is,
$16x^2-9y^2=576$
$\frac{16x^2}{576}-\frac{9y^2}{576}=1\Rightarrow \frac{x^2}{36}-\frac{y^2}{64}=1\text{Which is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The foci and vertices of the hyperbola lie on x - axis.
$\therefore a^2=36\Rightarrow a=6and{b}^2=64\Rightarrow b=8\text{Now}{c}^2=a^2+b^2=36+64=100\Rightarrow c=10\therefore \text{Coordinates of foci}(\pm c,0)i.e.(\pm 10,0)\text{Coordinates of vertices are}(\pm a,0)i.e.(\pm 6,0)\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}\text{Length of latus rectum}=\frac{2b^2}{a}=\frac{2\times 64}{6}=\frac{64}{3}.$