AB is a double ordinate of the hyperbola frac | vedclass

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(D) Let the coordinates of $A$ be $(x, l)$. Since $AB$ is a double ordinate,the coordinates of $B$ are $(x, -l)$.Given that $\Delta AOB$ is an equilat

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

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(D) Let the coordinates of $A$ be $(x, l)$. Since $AB$ is a double ordinate,the coordinates of $B$ are $(x, -l)$.Given that $\Delta AOB$ is an equilateral triangle,the side length $OA = OB = AB = 2l$.In $\Delta OMA$ (where $M$ is the midpoint of $AB$ on the $x$-axis),$OM = x$ and $AM = l$.Since $\angle AOM = 30^\circ$,we have $	an(30^\circ) = \frac{l}{x} = \frac{1}{\sqrt{3}}$,so $x = l\sqrt{3}$.Also,$OA^2 = x^2 + l^2 = (l\sqrt{3})^2 + l^2 = 3l^2 + l^2 = 4l^2$,which is consistent with $OA = 2l$.Since $A(x, l)$ lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $\frac{3l^2}{a^2} - \frac{l^2}{b^2} = 1$.$l^2 \left( \frac{3}{a^2} - \frac{1}{b^2} ight) = 1 \Rightarrow l^2 \left( \frac{3b^2 - a^2}{a^2b^2} ight) = 1$.Since $l^2 > 0$,we must have $3b^2 - a^2 > 0$,which implies $3b^2 > a^2$,or $\frac{b^2}{a^2} > \frac{1}{3}$.Using the relation $e^2 = 1 + \frac{b^2}{a^2}$,we get $e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.Therefore,$e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

$AB$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\Delta AOB$ (where $O$ is the origin) is an equilateral triangle. Then the eccentricity $e$ of the hyperbola satisfies:

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