Question

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

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

Answer 1

The correct option is D

$2$

Explanation for the correct option:

Finding the eccentricity:

We know that latus rectum of a hyperbola is given by $\frac{2b^2}{a}$ and semi conjugate axis is given by: $b$.

$$\begin{gathered} \therefore \frac{2b^2}{a}=12\left[\mathrm{Given}\right] \\ \Rightarrow \frac{b^2}{a}=6 \end{gathered}$$

Since, $b=2\sqrt{3}$ [Given]

$$\begin{gathered} \Rightarrow \frac{{\left(2\sqrt{3}\right)}^2}{a}=6 \\ \Rightarrow \frac{4\times 3}{a}=6 \\ \Rightarrow a=\frac{12}{6} \\ \Rightarrow a=2 \end{gathered}$$

So, $a^2=4$ and $b^2=12$.

The eccentricity of a hyperbola is given by

$\mathit{e}\mathbf{=}\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}}$

Substitute the values.

$$\begin{gathered} \therefore e=\sqrt{1+\frac{12}{4}} \\ =\sqrt{1+3} \\ =\sqrt{4} \\ =2 \end{gathered}$$

Therefore, the correct answer is option (D).