Question

Let $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{where}a>b>0,$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 $4\sqrt{3}.$

$\begin{array}{cccc}& LIST-I& & LIST-II\\ P.& \text{The length of the conjugate axis of}H\text{is}& 1.& 8\\ Q.& \text{The eccentricity of}H\text{is}& 2.& \frac{4}{\sqrt{3}}\\ R.& \text{The distance between the foci of}H\text{is}& 3.& \frac{2}{\sqrt{3}}\\ S.& \text{The length of the latus rectum of}H\text{is}& 4.& 4\end{array}$

The correct option is:

• A. $P\to 4;Q\to 2;R\to 1;S\to 3$
• B. $P\to 4;Q\to 3;R\to 1;S\to 2$
• C. $P\to 4;Q\to 1;R\to 3;S\to 2$
• D. $P\to 3;Q\to 4;R\to 2;S\to 1$

Answer 1

The correct option is B $P\to 4;Q\to 3;R\to 1;S\to 2$


$\tan {30}^{\circ }=\frac{b}{a}=\frac{1}{\sqrt{3}}$
$area\left(△LMN\right)=4\sqrt{3}=\frac{1}{2}\cdot \left(2b\right)\cdot a$
$\Rightarrow a\cdot b=a\left(\frac{a}{\sqrt{3}}\right)=4\sqrt{3}\Rightarrow a^2=12$
$\Rightarrow a=2\sqrt{3}$
$\therefore b=2$
Length of conjugate axis $2b=4$
$b^2=a^2(e^2-1)\Rightarrow \text{eccentricity}e=\frac{2}{\sqrt{3}}$
The distance between the foci of H $=2ae=2\cdot 2\sqrt{3}\cdot \frac{2}{\sqrt{3}}=8$
Length of latus rectum
$=\frac{2b^2}{a}=\frac{2\times 4}{2\sqrt{3}}=\frac{4}{\sqrt{3}}$