PQ is a double ordinate of the hyperbola frac | vedclass

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(D) Let the coordinates of $P$ be $(a \sec 	heta, b 	an 	heta)$. Since $PQ$ is a double ordinate, the coordinates of $Q$ are $(a \sec 	heta, -b

Vedclass · Accepted Answer

$e > 2 / \sqrt{3}$

Vedclass · Answer

(D) Let the coordinates of $P$ be $(a \sec 	heta, b 	an 	heta)$. Since $PQ$ is a double ordinate, the coordinates of $Q$ are $(a \sec 	heta, -b 	an 	heta)$.Given that $	riangle OPQ$ is an equilateral triangle, the angle $\angle POD = 30^{\circ}$, where $D$ is the point $(a \sec 	heta, 0)$.In $	riangle OPD$, we have $	an 30^{\circ} = \frac{PD}{OD} = \frac{b 	an 	heta}{a \sec 	heta}$.$\frac{1}{\sqrt{3}} = \frac{b}{a} \sin 	heta \implies \frac{b}{a} = \frac{1}{\sqrt{3} \sin 	heta}$.We know that $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{1}{3 \sin^2 	heta}$.Since $0  1$.Therefore, $e^2 = 1 + \frac{1}{3 \sin^2 	heta} > 1 + \frac{1}{3} = \frac{4}{3}$.Thus, $e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

$PQ$ 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 centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies:

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