Match Hyperbola Properties Based on Subtended Angle and Triangle Area

To determine the correct matching between List-I and List-II, we analyze the geometric properties of the given hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a > b > 0$.

1. Establishing equations for $a$ and $b$

• The endpoints of the conjugate axis $LM$ are $L(0, b)$ and $M(0, -b)$, giving the length of the conjugate axis as $LM = 2b$.
• A vertex of the hyperbola is $N(a, 0)$ (or $(-a, 0)$).
• The distance from the origin $O(0,0)$ to the vertex $N(a,0)$ is $ON = a$.

The conjugate axis $LM$ subtends an angle of $60^\circ$ at the vertex $N$. By symmetry across the $x$-axis, the line segment $ON$ bisects $\angle LNM$. Therefore, in the right-angled triangle $\triangle LON$: $\angle LNO = \frac{60^\circ}{2} = 30^\circ$

Using trigonometry in $\triangle LON$: $\tan(30^\circ) = \frac{OL}{ON} = \frac{b}{a}$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have: $\frac{b}{a} = \frac{1}{\sqrt{3}} \implies a = b\sqrt{3} \quad \text{--- (Equation 1)}$

The area of $\triangle LMN$ is given by: $\text{Area}(\triangle LMN) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (LM) \times (ON) = \frac{1}{2} \times (2b) \times a = ab$

Given that the area of $\triangle LMN$ is $4\sqrt{3}$: $ab = 4\sqrt{3} \quad \text{--- (Equation 2)}$

2. Solving for $a$ and $b$

Substitute $a = b\sqrt{3}$ from Equation (1) into Equation (2): $(b\sqrt{3}) \cdot b = 4\sqrt{3} \implies b^2 \sqrt{3} = 4\sqrt{3} \implies b^2 = 4 \implies b = 2$

Since $a = b\sqrt{3}$: $a = 2\sqrt{3}$

Notice that $a = 2\sqrt{3} \approx 3.464 > b = 2 > 0$, which satisfies the given condition $a > b > 0$.

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3. Evaluating the quantities in List-I

• P. Length of the conjugate axis: $\text{Length} = 2b = 2 \times 2 = 4$ Therefore, P $\rightarrow$ 4.

• Q. Eccentricity ($e$) of $H$: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$ Therefore, Q $\rightarrow$ 3.

• R. Distance between the foci: $\text{Distance} = 2ae = 2 \times (2\sqrt{3}) \times \frac{2}{\sqrt{3}} = 8$ Therefore, R $\rightarrow$ 1.

• S. Length of the latus rectum: $\text{Length} = \frac{2b^2}{a} = \frac{2 \times 4}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$ Therefore, S $\rightarrow$ 2.

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Conclusion

The correct matching is: $\text{P} \rightarrow 4; \quad \text{Q} \rightarrow 3; \quad \text{R} \rightarrow 1; \quad \text{S} \rightarrow 2$

This corresponds to Option B.