Question

If the length of the latus rectum and the length of the transverse axis of a hyperbola are $4\sqrt{3}$ and $2\sqrt{3}$ respectively, then the equation of the hyperbola is.

• A. $\frac{x^2}{3}-\frac{y^2}{4}=1$
• B. $\frac{x^2}{3}-\frac{y^2}{9}=1$
• C. $\frac{x^2}{6}-\frac{y^2}{4}=1$
• D. $\frac{x^2}{6}-\frac{y^2}{3}=1$
• E. $\frac{x^2}{3}-\frac{y^2}{6}=1$

Answer 1

The correct option is E

$\frac{x^2}{3}-\frac{y^2}{6}=1$

Explanation for the correct option:

Let us assume that the equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

The length of the transverse axis can be given by, $2a=2\sqrt{3}$.

$\Rightarrow a=\sqrt{3}$

The length of the latus rectum can be given by, $\frac{2b^2}{a}=4\sqrt{3}$.

Put $a=\sqrt{3}$ in the equation.

$$\begin{gathered} \Rightarrow \frac{2b^2}{\sqrt{3}}=4\sqrt{3} \\ \Rightarrow 2b^2=4\sqrt{3}\times \sqrt{3} \\ \Rightarrow 2b^2=12 \\ \Rightarrow b^2=6 \end{gathered}$$

Therefore, the equation of the hyperbola can be given by, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

Put $a=\sqrt{3}$ and $b^2=6$ in the above equation.

$$\begin{gathered} \Rightarrow \frac{x^2}{{\left(\sqrt{3}\right)}^2}-\frac{y^2}{6}=1 \\ \Rightarrow \frac{x^2}{3}-\frac{y^2}{6}=1 \end{gathered}$$

Hence, option (E) is the correct answer.