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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 $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..
| List $I$ | List $II$ |
| $P$ The length of the conjugate axis of $H$ is | $1$ $8$ |
| $Q$ The eccentricity of $H$ is | $2$ ${\frac{4}{\sqrt{3}}}$ |
| $R$ The distance between the foci of $H$ is | $3$ ${\frac{2}{\sqrt{3}}}$ |
| $S$ The length of the latus rectum of $H$ is | $4$ $4$ |
The correct option is:
$P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 3$
$P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 1 ; S \rightarrow 2$
$P \rightarrow 4 ; Q \rightarrow 1 ; R \rightarrow 3 ; S \rightarrow 2$
$P \rightarrow 3 ; Q \rightarrow 4 ; R \rightarrow 2 ; S \rightarrow 1$
Solution
$\triangle LMN =4 \sqrt{3}$
$\frac{1}{2} 2 ab =4 \sqrt{3}$
$ab =4 \sqrt{3}$
$\text { a. }\left(\frac{ a }{\sqrt{3}}\right)=4 \sqrt{3}$
$a ^2=12 \Rightarrow a =2 \sqrt{3}$
$\frac{ b }{ a }=\tan 30^{\circ}$
$b =\frac{2 \sqrt{3}}{\sqrt{3}}=20$
Length of conjugate axis $2 b =4$.
$e =\sqrt{1+\frac{4}{12}}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$
$SS ^{\prime}=2 ae =2 \times 2 \sqrt{3} \times \frac{2}{\sqrt{3}}=8$
$\text { L.R. }=\frac{2 b ^2}{ a }=\frac{2 \times 4}{2 \sqrt{3}}=\frac{4}{\sqrt{3}}$
