A tangent drawn to hyperbola (x2)/(a2) - (y2)/(b2) = 1 at P...

The point $P \left(\dfrac{\pi}{6}\right)is \$lleft(asec$\dfrac{\pi}{6}$, btan$\dfrac{\pi}{6}$)or$P \left(\dfrac{2a}{3}, \dfrac{b}{3}\right)

$∴$ Equation of tangent at $P$ is \dfrac{x}{\sqrt{3}a2} - \dfrac{y}{\sqrt{3}b} =1$1$

$∴$ Area of the triangle \text{ and } $= \times$\dfrac{\sqrt{}a}{} \times \sqrt{} b= 3a^{}$</p><p>$∴ \dfrac{b}{a} =$</p><p>$e^{} = +\dfrac{b}{a}=$<mn > 17$

Now, $e^{} - =<mn > 17 - =<mn > 8$

Hence, option (d) is correct.