Question

The asymptote of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ forms a triangle with any tangent to the hyperbola. The area of such triangle formed is $a^2\tan \lambda$ , where $\lambda \in (0,\frac{\pi }{2})$ then its eccentricity is

• A. $\sec \lambda$
• B. $\text{cosec}\lambda$
• C. ${\sec }^2\lambda$
• D. ${\text{cosec}}^2\lambda$

Answer 1

The correct option is A $\sec \lambda$

Any tangent to the hyperbola forms a triangle with the asymptotes which has constant area $ab$
$ab=a^2\tan \lambda \frac{b}{a}=\tan \lambda e^2=1+\frac{b^2}{a^2}=1+{\tan }^2\lambda ={\sec }^2\lambda e=\sec \lambda$

Note:
You don't need to know the property: "Any tangent to the hyperbola forms a triangle with the asymptotes which has constant area $ab$"
You can take a particular case, E.x. tangent at the vertex $(a,0)$ and proceed