Let H : frac{x^2}{a^2} - frac{y^2}{b^2} = 1,w | vedclass

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(A) The vertices of the conjugate axis are $L(0, b)$ and $M(0, -b)$. The vertex of the hyperbola is $N(a, 0)$.The length of the conjugate axis $LM = 2

Vedclass · Accepted Answer

$P$ $\rightarrow 4; Q$ $\rightarrow 3; R$ $\rightarrow 1; S$ $\rightarrow 2$

Vedclass · Answer

(A) The vertices of the conjugate axis are $L(0, b)$ and $M(0, -b)$. The vertex of the hyperbola is $N(a, 0)$.The length of the conjugate axis $LM = 2b$.The area of $	riangle LMN = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2b) 	imes a = ab = 4\sqrt{3}$.Since $\angle LNM = 60^{\circ}$,the angle $\angle LNO = 30^{\circ}$ (where $O$ is the origin).In $	riangle LNO$,$	an 30^{\circ} = \frac{OL}{ON} = \frac{b}{a} = \frac{1}{\sqrt{3}}$,so $a = b\sqrt{3}$.Substituting $a = b\sqrt{3}$ into $ab = 4\sqrt{3}$,we get $b(b\sqrt{3}) = 4\sqrt{3}$ $\Rightarrow b^2 = 4$ $\Rightarrow b = 2$.Then $a = 2\sqrt{3}$.$P$. Length of conjugate axis $= 2b = 2(2) = 4$.$Q$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.$R$. Distance between foci $= 2ae = 2(2\sqrt{3})(\frac{2}{\sqrt{3}}) = 8$.$S$. Length of latus rectum $= \frac{2b^2}{a} = \frac{2(4)}{2\sqrt{3}} = \frac{4}{\sqrt{3}}$.Thus,$P$ $ightarrow 4, Q$ $ightarrow 3, R$ $ightarrow 1, S$ $ightarrow 2$.

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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