Question

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

• 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 $P\to 4;Q\to 3;R\to 1;S\to 2$

$\tan {30}^{\circ }=\frac{b}{a}$

$\Rightarrow a=b\sqrt{3}$

Now area of $△LMN=\frac{1}{2}.2b.b\sqrt{3}$
$4\sqrt{3}=\sqrt{3}{b}^2$

$\Rightarrow b=2$ and $a=2\sqrt{3}$

$\Rightarrow e=\sqrt{1+\frac{b^2}{a^2}}=\frac{2}{\sqrt{3}}$

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 $=\frac{2b^2}{a}=\frac{2(2{)}^2}{2\sqrt{3}}=\frac{4}{\sqrt{3}}$
So $S\to 2$.