Let H : x 2 a 2 - y 2 b 2 = 1 , where a b 0 , be a hyperbola in the x y -plane whose conjugate axis L M subtends an angle of 60 ° at one of its vertices N . Let the area of the triangle L M N be 4 3 . [2018]

Question

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ where $a > b > 0$ , be a hyperbola in the $x y$ -plane whose conjugate axis $L M$ subtends an angle of $60 °$ at one of its vertices $N$ . Let the area of the triangle $L M N$ be $4 \sqrt{3}$ . [2018]

	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:

Smart Achievers · Accepted Answer

(2)

$Area of ∆ L M N = 4 \sqrt{3} (given)$

$\Rightarrow \frac{1}{2} \times L M \times O N = 4 \sqrt{3}$ $\Rightarrow \frac{1}{2} (2 b) (\sqrt{3} b) = 4 \sqrt{3}$

$∴$ $b^{2} = 4$ $\Rightarrow b = 2$

So, length of the conjugate axis of hyperbola $= 2 b = 4$

Now, $\tan 30 ° = \frac{O L}{O N} = \frac{b}{a}$ $H$ $\Rightarrow a = 2 \sqrt{3}$

$∴$ $b^{2} = a^{2} (e^{2} - 1)$ $\Rightarrow 4 = 12 (e^{2} - 1)$ $\Rightarrow e^{2} = 1 + \frac{1}{3} = \frac{4}{3}$

$∴$ $The eccentricity of hyperbola$ $e = \frac{2}{\sqrt{3}}$

$The distance between the foci of hyperbola$ $= 2 a e$

$= 2 \times 2 \sqrt{3} \times \frac{2}{\sqrt{3}}$ $= 8$

$And length of latus rectum of hyperbola$

$= \frac{2 b^{2}}{a}$ $= \frac{2 \times 4}{2 \sqrt{3}}$ $= \frac{4}{\sqrt{3}}$

Solution

1 P → 4; Q → 2; R → 1; S → 3

2 P → 4; Q → 3; R → 1; S → 2

3 P → 4; Q → 1; R → 3; S → 2

4 P → 3; Q → 4; R → 2; S → 1

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