If PQ is a double ordinate of hyperbola frac{ | vedclass

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Question

(d) Let $P$ $(a\sec 	heta ,\,b	an 	heta );\,Q(a\sec 	heta ,\, - b	an 	heta )$ be end points of double ordinates

and $C(0,\,0)$, is the centre

Vedclass · Accepted Answer

$e > 2/\sqrt 3 $

Vedclass · Answer

(d) Let $P$ $(a\sec heta ,\,b an heta );\,Q(a\sec heta ,\, - b an heta )$ be end points of double ordinates and $C(0,\,0)$, is the centre of the hyperbola. Now $PQ = 2b\, an heta $ $CQ = CP = \sqrt {{a^2}{{\sec }^2} heta + {b^2}{{ an }^2} heta } $ Since $CQ = CP = PQ$, $ herefore \,4{b^2}{ an ^2} heta = {a^2}{\sec ^2} heta + {b^2}{ an ^2} heta $ ==> $3{b^2}{ an ^2} heta = {a^2}{\sec ^2} heta $ ==> $3{b^2}{\sin ^2} heta = {a^2}$ ==> $3{a^2}({e^2} - 1){\sin ^2} heta = {a^2}$ ==> $3({e^2} - 1){\sin ^2} heta = 1$ ==> $\frac{1}{{3({e^2} - 1)}} = {\sin ^2} heta < 1$, $ \Rightarrow $ $\frac{1}{{{e^2} - 1}} < 3$ $\Rightarrow \,\,{e^2} - 1 > \frac{1}{3}$ $ \Rightarrow \,\,{e^2} > \frac{4}{3}$ $ \Rightarrow \,e > \frac{2}{{\sqrt 3 }}$.

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

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