If PQ is a double ordinate of the hyperbola f | vedclass

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(D) Let the coordinates of $P$ be $(\alpha, \beta)$. Since $PQ$ is a double ordinate,$Q$ is $(\alpha, -\beta)$.$PQ = 2|\beta|$ and $OP = \sqrt{\alpha^

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

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(D) Let the coordinates of $P$ be $(\alpha, \beta)$. Since $PQ$ is a double ordinate,$Q$ is $(\alpha, -\beta)$.$PQ = 2|\beta|$ and $OP = \sqrt{\alpha^2 + \beta^2}$.Since $	riangle OPQ$ is equilateral,$OP = PQ$,so $OP^2 = PQ^2$.$\alpha^2 + \beta^2 = (2\beta)^2 = 4\beta^2 \Rightarrow \alpha^2 = 3\beta^2$.Since $P(\alpha, \beta)$ lies on the hyperbola,$\frac{\alpha^2}{a^2} - \frac{\beta^2}{b^2} = 1$.Substituting $\alpha^2 = 3\beta^2$,we get $\frac{3\beta^2}{a^2} - \frac{\beta^2}{b^2} = 1$.$\beta^2 \left( \frac{3}{a^2} - \frac{1}{b^2} ight) = 1$.Since $\beta^2 > 0$,we must have $\frac{3}{a^2} - \frac{1}{b^2} > 0$,which implies $\frac{3}{a^2} > \frac{1}{b^2} \Rightarrow \frac{b^2}{a^2} > \frac{1}{3}$.Using $b^2 = a^2(e^2 - 1)$,we get $e^2 - 1 > \frac{1}{3}$.$e^2 > \frac{4}{3} \Rightarrow e > \frac{2}{\sqrt{3}}$.

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the center of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

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