Let H:x2a2−y2b2=1 , where a>b>0be 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 43LIST-I LIST-II P. The length of the conjugate axis of H is 1. 8 Q. The eccentricity of H is 2. 43 R. The distance between the foci of H is 3. 23 S. The length of the latus rectum of H is 4. 4

tan30°=ba ⇒a=b3 Now are of ΔLMN=12..b.b3 43=3b2 ⇒b=2&a=23 P. Length of conjugate axis =2 b=4 So P→4 Q. Eccentricity e=23 So Q→3 R . distance between foci =2ae=2(23)23=8 So R→1 S. Length of latus rectum =2b2a=2(2)223=43 So S→2

Let H:x2a2−y2b2=1 , where a>b>0be 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 43LIST-I LIST-II P. The length of the conjugate axis of H is 1. 8 Q. The eccentricity of H is 2. 43 R. The distance between the foci of H is 3. 23 S. The length of the latus rectum of H is 4. 4

tan30°=ba ⇒a=b3 Now are of ΔLMN=12..b.b3 43=3b2 ⇒b=2&a=23 P. Length of conjugate axis =2 b=4 So P→4 Q. Eccentricity e=23 So Q→3 R . distance between foci =2ae=2(23)23=8 So R→1 S. Length of latus rectum =2b2a=2(2)223=43 So S→2

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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 $60 °$ 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

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P-3;Q-4;R-2;S-1

P-4;Q-2;R-1;S-3

P-4;Q-1;R-3;S-2

P-4;Q-3;R-1;S-2

answer is B.

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Detailed Solution

$\begin{array}{l} \tan 30 ° = \frac{b}{a} \Rightarrow a = b \sqrt{3} \\ Now are of Δ L M N = \frac{1}{2} . . b . b \sqrt{3} 4 \sqrt{3} = \sqrt{3} b^{2} \Rightarrow b = 2 & a = 2 \sqrt{3} \end{array}$