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$Angle between the asymptotes of a hyperbola is 30^{\circ} then e =$

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$\sqrt{6} - \sqrt{2}$

$\sqrt{6} - \sqrt{3}$

$\sqrt{6}$

$\sqrt{2}$

answer is C.

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Detailed Solution

$G i v e n a n g l e b e t w e e n a s y m p t o t e s θ = 30^{0} I f θ i s a n g l e b e t w e e n a s y m p t o t e s t h e n θ = 2 s e c^{- 1} e \Rightarrow e = s e c 15^{0} = \frac{2 \sqrt{2}}{\sqrt{3} + 1} = \sqrt{6} - \sqrt{2}$

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