A double ordinate PQ of the hyperbola frac{x^ | vedclass

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(D) Let the coordinates of $P$ be $(x_1, y_1)$. Since $PQ$ is a double ordinate, $Q$ is $(x_1, -y_1)$.Since $\Delta OPQ$ is equilateral and $M$ is the

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

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(D) Let the coordinates of $P$ be $(x_1, y_1)$. Since $PQ$ is a double ordinate, $Q$ is $(x_1, -y_1)$.Since $\Delta OPQ$ is equilateral and $M$ is the midpoint of $PQ$ on the $x$-axis, $\angle POM = 30^{\circ}$.In $\Delta OMP$, $	an 30^{\circ} = \frac{PM}{OM} = \frac{y_1}{x_1} = \frac{1}{\sqrt{3}}$.Thus, $x_1 = \sqrt{3} y_1$.Since $P(x_1, y_1)$ lies on the hyperbola, $\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$.Substituting $x_1 = \sqrt{3} y_1$, we get $\frac{3y_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$, which implies $y_1^2 \left( \frac{3b^2 - a^2}{a^2b^2} ight) = 1$.For $y_1^2 > 0$, we must have $3b^2 - a^2 > 0$, so $b^2 > \frac{a^2}{3}$, or $\frac{b^2}{a^2} > \frac{1}{3}$.Using $e^2 = 1 + \frac{b^2}{a^2}$, we have $e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.Therefore, $e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

$A$ double ordinate $PQ$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is such that $\Delta OPQ$ is equilateral, where $O$ is the centre of the hyperbola. Then the eccentricity $e$ satisfies the relation:

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