Equation and key points of the hyperbola

Given $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \qquad a>b>0.$

For this hyperbola:

• Vertices are $N=(\pm a,0)$
• Conjugate axis endpoints are $L=(0,b)$ and $M=(0,-b)$
• Length of conjugate axis $=2b$
• Eccentricity $e=\frac{c}{a}$ where $c^2=a^2+b^2$
• Distance between foci $=2c$
• Length of latus rectum $=\dfrac{2b^2}{a}$

Use the angle condition

The conjugate axis $LM$ subtends an angle $60^\circ$ at one vertex $N$.

Take the vertex $N=(a,0)$. Then the angle $\angle LNM=60^\circ$.

Vectors: $\overrightarrow{NL}=(-a,b), \qquad \overrightarrow{NM}=(-a,-b).$

Using the dot product, \cos \angle LNM=rac{\overrightarrow{NL}\cdot\overrightarrow{NM}}{|\overrightarrow{NL}|\,|\overrightarrow{NM}|}.

Now, $\overrightarrow{NL}\cdot\overrightarrow{NM}=a^2-b^2,$ and $|\overrightarrow{NL}|=|\overrightarrow{NM}|=\sqrt{a^2+b^2}.$

So, $\cos 60^\circ=\frac{a^2-b^2}{a^2+b^2}.$

Since $\cos 60^\circ=\frac12$, $\frac{a^2-b^2}{a^2+b^2}=\frac12.$

Therefore, $2a^2-2b^2=a^2+b^2$ $a^2=3b^2.$

Use the area condition

Triangle $LMN$ has base $LM=2b$ and the perpendicular distance from $N=(a,0)$ to the line $x=0$ is $a.$

Hence, $\text{Area}(\triangle LMN)=\frac12\cdot 2b\cdot a=ab.$

Given area is $4\sqrt3$, so $ab=4\sqrt3.$

From $a^2=3b^2$, since $a>b>0$, $a=\sqrt3\,b.$

Substitute into $ab=4\sqrt3$: $\sqrt3\,b^2=4\sqrt3$ $b^2=4$ $b=2.$

Then $a=\sqrt3\cdot 2=2\sqrt3.$

Now compute the required quantities

P. Length of conjugate axis

$2b=2(2)=4.$ So, $P \to 4.$ In List-II, this is 4.

Q. Eccentricity

$c^2=a^2+b^2=(2\sqrt3)^2+2^2=12+4=16$ $c=4.$

Therefore, $e=\frac{c}{a}=\frac{4}{2\sqrt3}=\frac{2}{\sqrt3}.$ So, $Q \to \frac{2}{\sqrt3}.$ In List-II, this is 3.

R. Distance between the foci

$2c=2(4)=8.$ So, $R \to 8.$ In List-II, this is 1.

S. Length of latus rectum

$\frac{2b^2}{a}=\frac{2\cdot 4}{2\sqrt3}=\frac{4}{\sqrt3}.$ So, $S \to \frac{4}{\sqrt3}.$ In List-II, this is 2.

Final matching

Thus,

• $P \to 4$
• $Q \to 3$
• $R \to 1$
• $S \to 2$

This corresponds to Option B.