A tangent drawn to the hyperbola x^ 2 a^ 2 - y^ 2 b^ 2 = 1 at P (...

$P = \left(\dfrac{2a}{\sqrt{3}}, \dfrac{b}{\sqrt{3}}\right)$
Equation of the tangent at $P$ is $\dfrac{2x}{\sqrt{3}a} - \dfrac{y}{\sqrt{3}b} = 1$
$x$ intercept & $y$ intercept of the tangent are $\dfrac{\sqrt{3}a}{2}$ & $- \sqrt{3}b$ respectively
Area of the triangle formed by the tangent with the coordinate axes is
$\dfrac{1}{2} \times \dfrac{\sqrt{3}a}{2} \times \sqrt{3}b = 3{a}^{2}$
$⇒ \dfrac{b}{a} = 4 ⇒ \dfrac{a}{b} = \dfrac{1}{4}$
The eccentricity of the conjugate hyperbola equals to
$\sqrt{1 + \dfrac{1}{16}} = \dfrac{\sqrt{17}}{4}$