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Q.

AB is double ordinate of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1,$ such that $△ A O B$ (o is origin) is an equilateral triangle, then eccentricity e satisfies-----

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a

$e > \sqrt{3}$

b

$1 < e < \frac{2}{\sqrt{3}}$

c

$e = \frac{2}{\sqrt{3}}$

d

$e > \frac{2}{\sqrt{3}}$

answer is D.

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Detailed Solution

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$p u t A = (\frac{l \sqrt{3}}{2}, \frac{l}{2})$ in hyperbola equation,

We get $l^{2} (\frac{3}{4 a^{2}} - \frac{1}{4 b^{2}}) = 1$

$\Rightarrow \frac{3}{4 a^{2}} - \frac{1}{4 b^{2}} = \frac{1}{l^{2}} > 0 \Rightarrow \frac{b^{2}}{a^{2}} > \frac{1}{3}$

$\Rightarrow e^{2} - 1 > \frac{1}{3}$

$\Rightarrow e^{2} > \frac{4}{3} (i e) e > \frac{2}{\sqrt{3}} .$

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