The asymptotes of the hyperbola frac{x^2}{a^2 | vedclass

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(A) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The asymptotes of the hyperbola are given by $\frac{x}{a} \pm \frac{y}{b} =

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$\sec (\alpha)$

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(A) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The asymptotes of the hyperbola are given by $\frac{x}{a} \pm \frac{y}{b} = 0$. It is a standard property that the area of the triangle formed by any tangent to the hyperbola and its asymptotes is constant and equal to $ab$. Given that the area is $a^2 	an(\alpha)$,we have $ab = a^2 	an(\alpha)$,which implies $b = a 	an(\alpha)$. The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substituting $b = a 	an(\alpha)$,we get $e = \sqrt{1 + \frac{a^2 	an^2(\alpha)}{a^2}} = \sqrt{1 + 	an^2(\alpha)} = \sqrt{\sec^2(\alpha)} = \sec(\alpha)$.

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 asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,with any tangent to the hyperbola form a triangle whose area is $a^2 \tan (\alpha)$. Then its eccentricity equals

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