Question

The latusrectum of the hyperbola $\frac{x^2}{4}-\frac{y^2}{12}=1$ is 4 $\frac{1}{2}$ . Its eccentricity e =

• A. 4/5
• B. 5/4
• C. 3/4
• D. 4/3

Answer 1

The correct option is B

5/4

$\frac{x^2}{4}-\frac{y^2}{12}$ = 1 $\frac{2b^2}{a}=4\frac{1}{2}=\frac{9}{2}$

a = 4 $\Rightarrow$ $4b^2=9a$

$\Rightarrow$ $4a^2(e^2-1)=9a$

$\Rightarrow$ $e^2-1=\frac{9}{16}$

$\Rightarrow$ $e^2=\frac{25}{16}$

$\Rightarrow$ e = $\frac{5}{4}$