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

As we know that any tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ forms a triangle with the asymptotes which has constant area $ab$.

$\text{ Given, } ab=a^2\tan \lambda$

 $\text{⇒ }b=a\tan \lambda$

$\Rightarrow b^2=a^2{\tan }^2\lambda$

Since $b^2=a^2\left(e^2-1\right)$

$\Rightarrow a^2\left(e^2-1\right)=a^2{\tan }^2\lambda$

$\Rightarrow e^2=1+{\tan }^2\lambda ={\sec }^2\lambda$

$\Rightarrow e=\sec \lambda$

Hence option-$1$ is the correct answer.