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

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Solution

The correct option is D
e >

Let the length of the double ordinate be $2^{l}$

$∴ A B = 2^{l} a n d A M = B M = l$

Clearly ordinate of point A is $l$

The abscissa of the point A is given by

$\frac{x^{2}}{a^{2}} - \frac{l^{2}}{b^{2}} = 1 \Rightarrow x = \frac{a \sqrt{b^{2} + l^{2}}}{b}$

$∴ A i s (\frac{a \sqrt{b^{2} + l^{2}}}{b}, l)$

Since $Δ O A B$ is equilateral triangle, therefore

$O A = A B = O B = 2^{l}$

$A l s o, O M^{2} + A M^{2} = O A^{2} ∴ \frac{a (b^{2} + l^{2})}{b} + l^{2} = 4 l^{2}$

$W e g e t l^{2} = \frac{a^{2} b^{2}}{3 b^{2} - a^{2}}$

$S i n c e l^{2}$ > 0 $∴ \frac{a^{2} b^{2}}{3 b^{2} - a^{2}}$ > $0 \Rightarrow 3 b^{2} - a^{2}$ >0

$\Rightarrow 3 a^{2} (e^{2} - 1) - a^{2}$ > $0 \Rightarrow e$ > $\frac{2}{\sqrt{3}}$

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If AB is a double ordinate of the hyperbola such that ∆ OAB (O is the origin) is an equilateral triangle, then the eccentricity e of the hyperbola satisfies

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