AB is double ordinate of the hyperbola `x^2/a^2-y^2/b^2=1` such that `DeltaAOB`(where 'O' is the origin) is an equilateral triangle, then the eccentricity e of hyperbola satisfies:

Options

A. `1 lt e lt (2)/(sqrt(3))`
B. `e=(2)/(sqrt(3))`
C. `e=(sqrt(3))/(2)`
D. `e lt(2)/(sqrt(3))`

Answer

D

Explanation

Let the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` and any double ordinate PQ be (a `sec theta, b tan theta`),(`a sec theta, - b tan theta)`.
O is the centre(0,0).
`Delta OPQ` is euilateral. So
`tan 30^(@)=(b,tan theta)/(a sec theta)`
`rArr3(b^(2))/(a^(2))=cosec^(2)theta`
`rArr3(e^(2)-1)=cosec^(2)theta`
Now `cosec^(2)theta ge 1`
`rArr 3 (e^(2)-1)ge 1 " " rArr e^(2)ge (4)/(3) rArr e gt (2)/(sqrt(3))`