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 equation of given hyperbola is,
$\frac{x^2}{16}-\frac{y^2}{9}=1\text{Which is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{The foci and vertices of the hyperbola lie on x - axis}.\therefore a^2=16\Rightarrow a=4and{b}^2=9\Rightarrow b=3\text{Now}{c}^2=a^2+b^2=16+=25\Rightarrow c=5\therefore \text{Coordinates fo foci are}(\pm c,0)i.e.(\pm 5,0)\text{coordinates of vertices are}(\pm a,0)i.e.(\pm 4,0)\text{Eccentricity}\left(e\right)=\frac{c}{a}=\frac{5}{4}\text{Length of latus rectum}=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}$