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

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(D) Given the equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $\Delta OPQ$ is an equilateral triangle. $PQ$ is a double o

Vedclass · Accepted Answer

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

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(D) Given the equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $\Delta OPQ$ is an equilateral triangle. $PQ$ is a double ordinate of the hyperbola.Let the coordinates of $P$ and $Q$ be $(a \sec 	heta, b 	an 	heta)$ and $(a \sec 	heta, -b 	an 	heta)$ respectively.In $\Delta OPQ$, $OP = PQ$.Since $\Delta OPQ$ is equilateral, $OP^2 = PQ^2$.$OP^2 = (a \sec 	heta)^2 + (b 	an 	heta)^2 = a^2 \sec^2 	heta + b^2 	an^2 	heta$.$PQ^2 = (2b 	an 	heta)^2 = 4b^2 	an^2 	heta$.Equating them: $a^2 \sec^2 	heta + b^2 	an^2 	heta = 4b^2 	an^2 	heta$.$a^2 \sec^2 	heta = 3b^2 	an^2 	heta$.$a^2 (1 + 	an^2 	heta) = 3b^2 	an^2 	heta$.$a^2 = (3b^2 - a^2) 	an^2 	heta$.$	an^2 	heta = \frac{a^2}{3b^2 - a^2}$.Since $	an^2 	heta > 0$, we must have $3b^2 - a^2 > 0$, so $3b^2 > a^2$, or $\frac{b^2}{a^2} > \frac{1}{3}$.We know $e^2 = 1 + \frac{b^2}{a^2}$.Since $\frac{b^2}{a^2} > \frac{1}{3}$, $e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.Therefore, $e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

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

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