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Question:

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

P→4; Q→2; R→1; S→3

P→4; Q→1; R→3; S→2

P→4; Q→3; R→1; S→2

P→3; Q→4; R→2; S→1

Solution:

tan 30° = b/a ⇒ a = b√3
Now area of ΔLMN = 1/2. 2b. b√3 / 4√3 = √3b²/2 ⇒ b = 2 and a = 2√3 ⇒ e = √(1 + b²/a²) = 2√3
P. Length of conjugate axis = 2b = 4 So P→4
Q. Eccentricity e = 2√3 So Q→3
R. Distance between foci = 2ae = 2(2√3)(2√3) = 8 So R→1
S. Length of latus rectum = 2b²/a = 2(2)²/2√3 = 4/√3 So S→2