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  • A B  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%20%5Cfrac%7Bx%5E2%7D%7B

A B  is a double ordinate of the hyperbola \mathrm{ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1} such that  \mathrm{ \triangle A O B} (where\mathrm{ ' O^{\prime} }is the origin)  is an equilateral triangle, then the eccentricity e of the hyperbola satisfies

Option: 1

\mathrm{e>\sqrt{3}}


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}         \mathrm{\text { where } \mathrm{y}=l}

\mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}=1+\frac{l^2}{\mathrm{~b}^2} \quad \Rightarrow \mathrm{x}^2=\left(\mathrm{b}^2+l^2\right) \frac{\mathrm{a}^2}{\mathrm{~b}^2}}          ....(1)              

\mathrm{\text { now } \mathrm{x}^2+R^2=4 l^2 \Rightarrow \mathrm{x}^2=3 l^2}                        ....(2)

from (1) and (2)  \mathrm{\frac{\mathrm{a}^2\left(\mathrm{~b}^2+l^2\right)}{\mathrm{b}^2}=3 l^2 \Rightarrow \mathrm{a}^2 \mathrm{~b}^2+\mathrm{a}^2 l^2=3 \mathrm{~b}^2 l^2}

\mathrm{l^2\left(3 \mathrm{~b}^2-\mathrm{a}^2\right)=\mathrm{a}^2 \mathrm{~b}^2}

\mathrm{l^2=\frac{\mathrm{a}^2 \mathrm{~b}^2}{3 \mathrm{~b}^2-\mathrm{a}^2}>0 \quad \Rightarrow \quad 3 \mathrm{~b}^2-\mathrm{a}^2>0 \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}>\frac{1}{3}}

\mathrm{1+\frac{b^2}{a^2}>\frac{4}{3} \Rightarrow e^2>\frac{4}{3} \Rightarrow e>\frac{2}{\sqrt{3}}}

Posted by

Devendra Khairwa

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