Area of the triangle formed by the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and any tangent to the hyperbola is $a^2\tan\lambda$ in magnitude then its eccentricity is:
Step-by-Step Solution
Key Concept: The eccentricity of a hyperbola relates to the angle its asymptotes make through $e = \sec \lambda$.
Any tangent to a hyperbola forms a triangle with its asymptotes having constant area $ab$. Given $\frac{b}{a} = \tan \lambda$, we derive $e^2 - 1 = \tan^2 \lambda$, which yields $e^2 = 1 + \tan^2 \lambda = \sec^2 \lambda$, therefore $e = \sec \lambda$.
Correct Answer: 1