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AB is double ordinate of the hyperbola x...

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:

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With the given details. we can draw a hyperbola.
Please refer to video for the diagram.
Now, coordinates of point `B` are `(asec theta, btan theta)`.
As angle `AOB = 60^@`
`:.` Slope `= tan30^@ = (btan theta)/(a sectheta)`
`=>bsin theta/a = 1/sqrt3`
`=>b/a = 1/(sqrt3sin theta)`
Now, we know,
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