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The asymptote of the hyperbola x^2/a^2-y...

The asymptote of the hyperbola `x^2/a^2-y^2/b^2=1` form with an tangent to the hyperbola triangle whose area is `a^2 tan lambda` in magnitude then its eccentricity is: (a) `sec lambda` (b) `cosec lambda` (c) `sec^2 lambda` (d) `cosec^2 lambda`

Text Solution

Verified by Experts

The correct Answer is:
`e=sec lambda`
Any tangent to the hyperbola forms a triangle with the asymptotes which has constant area ab.
Given
`ab=a^(2) tan lambda`
`"or "(b)/(a)=tan lambda`
`"or "e^(2)-1=tan^(2) lambda`
`"or "e^(2)=1+tan^(2)lambda=sec^(2)lambda`
`"or "e=sec lambda`
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