Question

# The asymptotes of the hyperbola $$\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$$ form with any tangent to the hyperbola a triangle whose area is $$a^2 \tan\lambda$$ in magnitude, then its eccentricity is :

A
secλ
B
cosecλ
C
sec2λ
D
cosec2λ

Solution

## The correct option is A $$\sec \lambda$$Any tangent to hyperbola forms a triangle with the asymptotes which has constant area $$ab$$.$$\Rightarrow ab=a^2 \tan\lambda$$$$\displaystyle \Rightarrow \frac {b}{a}=\tan \lambda$$$$\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}$$$$\Rightarrow e = \sqrt{1+\tan^2{\lambda}} =\sec{\lambda}$$Maths

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