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  • If PQ is a double ordinate of the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5

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

 

Option: 1

\mathrm{1<e<\frac{2}{\sqrt{3}}}


Option: 2

\mathrm{e=\frac{2}{\sqrt{3}}}


Option: 3

\mathrm{e=\frac{\sqrt{3}}{2}}


Option: 4

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


Answers (1)

best_answer

\mathrm{\text { Let } P(a \sec \theta, b \tan \theta), Q(a \sec \theta,-b \tan \theta)} be end points of double oridinate and C(0, 0) is the centre of the hyperbola.

\mathrm{\text { Now, } P Q=2 b \tan \theta}

\mathrm{C Q=C P=\sqrt{a^2 \sec ^2 \theta+b^2 \tan ^2 \theta}}

\mathrm{\text { Since } C Q=C P=P Q}

\mathrm{\begin{aligned} & \therefore \quad 4 b^2 \tan ^2 \theta=a^2 \sec ^2 \theta+b^2 \tan ^2 \theta \\ & \Rightarrow \quad 3 b^2 \tan ^2 \theta=a^2 \sec ^2 \theta \Rightarrow 3 b^2 \sin ^2 \theta=a^2 \\ & \Rightarrow \quad 3 a^2\left(e^2-1\right) \sin ^2 \theta=a^2 \Rightarrow 3\left(e^2-1\right) \sin ^2 \theta=1 \\ & \Rightarrow \quad \frac{1}{3\left(e^2-1\right)}=\sin ^2 \theta<1 \quad\left(\quad \sin ^2 \theta<1\right) \\ & \Rightarrow \quad \frac{1}{e^2-1}<3 \Rightarrow e^2-1>\frac{1}{3} \quad \Rightarrow \quad e^2>\frac{4}{3} \Rightarrow e>\frac{2}{\sqrt{3}} \end{aligned}}

 

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Rishabh

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