Let $H:\frac{x^2}{a^2 }-\frac{y^2}{b^2}=1$ , where $a>b>0$, be a hyperbola in the xy-plane whose conjugate axis $LM$ subtends an angle of ${60}^o$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4\sqrt{3}$.

LIST-I	LIST-II
A. The length of the conjugate axis of $H$ is	P. $8$
B. The eccentricity of $H$ is	Q. $\frac{4}{\sqrt{3}}$
C. The distance between the foci of $H$ is	R. $\frac{2}{\sqrt{3}}$
D. The length of the latus rectum of $H$ is	S. $4$

The correct option is:

• A

  a-r;b-q;c-s;d-p;
• B

  a-q;b-r;c-s;d-p;
• C

  a-r;b-s;c-p;d-q;
• Correct answer
  D

  a-s;b-r;c-p;d-q;

Solution

$\tan {30}^o=\frac{b}{a}$
$\Rightarrow$ $a=b\sqrt{3}$
Now area of $∆LMN=\frac{1}{2}.2b.b\sqrt{3}$
$4\sqrt{3}=\sqrt{3}{b}^2$
$\Rightarrow b=2 \& a=2\sqrt{3}$
$\Rightarrow e=\sqrt{1+\frac{b^2}{a^2}}=\frac{2}{\sqrt{3}}$
P. Length of conjugate axis $=2b=4$
So $P\to 4$
Q. Eccentricity $e=\frac{2}{\sqrt{3}}$
So $Q\to 3$
R. Distance between foci $=2ae$
$=2\left(2\sqrt{3}\right)\left(\frac{2}{\sqrt{3}}\right)=8$
So, $R\to 1$
S. Length of latus rectum $=\frac{2b^2}{a}=\frac{2{\left(2\right)}^2}{2\sqrt{3}}=\frac{4}{\sqrt{3}}$
So $S\to 2$

Hence, the correct answer is option: 'D'.