Question

The latus rectum of the hyperbola $\frac{x^2}{16}-\frac{y^2}{b}=1$ is $4\frac{1}{2}$. Its eccentricity $e=$

• A. $\frac{4}{5}$
• B. $\frac{5}{4}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{5}{4}$

Length of latus rectum $=\frac{2b^2}{a}=\frac{9}{2}$
$\frac{2b^2}{4}=\frac{9}{2}\Rightarrow b^2=9$
$\Rightarrow b=3$
$e^2=\frac{a^2+b^2}{a^2}=\frac{16+9}{16}=\frac{25}{16}$
$e=\frac{5}{4}$