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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
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B
cosecλ
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C
sec2λ
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D
cosec2λ
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Solution

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

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