A tangent drawn to hyperbola x 2 a 2 - y 2 b 2 = 1 at P π 6 forms a triangle of area 3 a 2 square units…

The point $P\left(\frac{\pi }{6}\right)$ is $\left(a\sec \frac{\pi }{6},b\tan \frac{\pi }{6}\right)$ or $P\left(\frac{2a}{\sqrt{3}},\frac{b}{\sqrt{3}}\right)$

$\therefore$ Equation of tangent at $P$ is $\frac{x}{\frac{\sqrt{3}a}{2}}-\frac{y}{\sqrt{3}b}=1$1$

$\therefore$ Area of the triangle $=\frac{1}{2}\times \frac{\sqrt{3}a}{2}\times \sqrt{3}b=3a^2$

$\therefore \frac{b}{a}=4$

$e^2=1+\frac{b^2}{a^2}=17$

Now, $e^2-9=17-9=8$

Hence, option (d) is correct.