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

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(D) Let $P(a \sec 	heta, b 	an 	heta)$ and $Q(a \sec 	heta, -b 	an 	heta)$ be the endpoints of the double ordinate.$O(0, 0)$ is the centre of th

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$e > 2/\sqrt{3}$

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(D) Let $P(a \sec 	heta, b 	an 	heta)$ and $Q(a \sec 	heta, -b 	an 	heta)$ be the endpoints of the double ordinate.$O(0, 0)$ is the centre of the hyperbola.The length of the double ordinate $PQ = 2b 	an 	heta$.Since $	riangle OPQ$ is an equilateral triangle,$OP = OQ = PQ$.$OP^2 = (a \sec 	heta)^2 + (b 	an 	heta)^2 = a^2 \sec^2 	heta + b^2 	an^2 	heta$.Also,$PQ^2 = (2b 	an 	heta)^2 = 4b^2 	an^2 	heta$.Equating $OP^2 = PQ^2$,we get $a^2 \sec^2 	heta + b^2 	an^2 	heta = 4b^2 	an^2 	heta$.$a^2 \sec^2 	heta = 3b^2 	an^2 	heta$.$a^2 \frac{1}{\cos^2 	heta} = 3b^2 \frac{\sin^2 	heta}{\cos^2 	heta} \implies a^2 = 3b^2 \sin^2 	heta$.Using $b^2 = a^2(e^2 - 1)$,we have $a^2 = 3a^2(e^2 - 1) \sin^2 	heta$.$1 = 3(e^2 - 1) \sin^2 	heta \implies \sin^2 	heta = \frac{1}{3(e^2 - 1)}$.Since $0 $3(e^2 - 1) > 1 \implies e^2 - 1 > 1/3 \implies e^2 > 4/3$.Thus,$e > 2/\sqrt{3}$.

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 centre of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

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