Gujarati
Hindi
10-2. Parabola, Ellipse, Hyperbola
normal

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

A

$\sec \lambda$

B

$ cosec\lambda$

C

$\sec^2\lambda$

D

$cosec^2\lambda$

Solution

$A = ab = a^2 \tan \lambda$

$ \Rightarrow $  $b/a = \tan \lambda ,$

hence $e^2 = 1 + (b^2/a^2) $

$\Rightarrow $ $e^2 = 1 + \tan^2 \lambda$

$ \Rightarrow$  $ e= sec \lambda$   

Standard 11
Mathematics

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