Question

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

Answer 1

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 $a^2+b^2=c^2,$ where $c=ae$
$\therefore c^2=4^2+3^2=25$
$\Rightarrow c=5$
Therefore, the coordinates of the foci are $(\pm 5,0)$
The coordinates of the vertices are $(\pm 4,0)$
Eccentricity $e=$ $\frac{c}{a}=\frac{5}{4}$
Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}$