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  • If is a double ordinate of hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src

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

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Let \mathrm{P(a \sec \theta, b \tan \theta)\ ;\ Q(a \sec \theta,-b \tan \theta)} be end points of double ordinates and \mathrm{C(0,0)} is the centre of the hyperbola
\mathrm{Now\ P Q=2 b \tan \theta\ ; \quad C Q=C P=\sqrt{a^2 \sec ^2 \theta+b^2 \tan ^2 \theta}}\\\ \mathrm{Since\ C Q=C P=P Q, \quad \therefore \quad 4 b^2 \tan ^2 \theta=a^2 \sec ^2 \theta+b^2 \tan ^2 \theta}
\mathrm{\begin{aligned} &\mathrm{ \Rightarrow 3 b^2 \tan ^2 \theta=a^2 \sec ^2 \theta \Rightarrow 3 b^2 \sin ^2 \theta=a^2} \\ &\mathrm{ \Rightarrow 3 a^2\left(e^2-1\right) \sin ^2 \theta=a^2 \Rightarrow 3\left(e^2-1\right) \sin ^2 \theta=1} \end{aligned}}
\mathrm{\Rightarrow \quad \frac{1}{3\left(e^2-1\right)}=\sin ^2 \theta<1 \quad\left(\sin ^2 \theta<1\right)}
\mathrm{\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}}} 
   

Posted by

Devendra Khairwa

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