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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 $a^{2} \tan (a)$ then its eccentricity equals

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$\sec (a)$

$\cos e c (a)$

$\sec^{2} (a)$

$\cos e c^{2} (a)$

answer is A.

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Detailed Solution

Any tangent to hyperbola forms a triangle with asymptotes which has constant area = ab
$a^{2} \tan (a) = a b \frac{b}{a} = \tan (a) e = \sqrt{\frac{a^{2} + b^{2}}{a^{2}}} = \sqrt{1 + {(\frac{b}{a})}^{2}} = \sqrt{1 + \tan^{2} (a)} = \sec (a)$

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