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:

Let H : x²/a² - y²/b² = 1, where a > b > 0, be a hyperbola in the xy-plane whose conjugate axis LM subtends an angle of 60^° at one of its vertices N. Let the area of the triangle LMN be 4√3. | List I | List II | | --- | --- | | (P) The length of the conjugate axis of H is | (1) 8 | | (Q) The eccentricity of H is | (2) 4/√3 | | (R) The distance between the foci of H is | (3) 2/√3 | | (S) The length of the latus rectum of H is | (4) 4 | The correct option is:

$H: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,\quad a>b>0$

$\text{conjugate axis } LM \text{ subtends } 60^\circ \text{ at vertex } N$

Area of triangle LMN = 4√3

$\text{To find: conjugate axis, } e,\ \text{focal distance, latus rectum}$

$H:\ \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 \ (\text{transverse axis}=x\text{-axis})$

$\text{vertices } (\pm a,0);\quad \text{conjugate axis } LM:\ (0,\pm b)$

$\text{length of conjugate axis}=2b$

$N=(a,0)$

∠ LNM = 60^∘, so ∠ LNO = ∠ MNO = 30^∘

$\text{Plan: area} \Rightarrow ab;\quad \text{angle} \Rightarrow \text{relation } a,b$

$\text{then find } 2b,\ e,\ 2c,\ \dfrac{2b^2}{a}$

$\text{Area}=\dfrac{1}{2}(2b)(a)=ab=4\sqrt{3}$

$\tan 30^\circ = \dfrac{LO}{ON} = \dfrac{b}{a}$

$\dfrac{b}{a}=\dfrac{1}{\sqrt{3}} \Rightarrow a=\sqrt{3}\,b \Rightarrow a^2=3b^2$

$(\sqrt{3}\,b)\,b = 4\sqrt{3} \Rightarrow b^2=4 \Rightarrow b=2$

$a=\sqrt{3}\cdot 2 = 2\sqrt{3}$

$(P)\ 2b = 2(2) = 4 \to (4)$

$(Q)\ e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{4}{12}}=\dfrac{2}{\sqrt{3}} \to (3)$

$(R)\ c=\sqrt{a^2+b^2}=\sqrt{12+4}=4 \Rightarrow 2c=8 \to (1)$

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

$P\to 4,\ Q\to 3,\ R\to 1,\ S\to 2 \Rightarrow \text{option } (2)$