Question

The eccentricity of the hyperbola with latusrectum $12$ and semi-conjugate axis $2\sqrt{3}$ is

• A. $2$
• B. $3$
• C. $\frac{\sqrt{3}}{2}$
• D. $2\sqrt{3}$

Answer 1

Answer: (A)

$2$

We have $\frac{2b^2}{a}=12,b=2\sqrt{3}$

$\Rightarrow a=\frac{8\times 3}{12}$ $=2$
Therefore, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{12}{4}}=2$