Given $\frac{2 \mathrm{~b}^2}{\mathrm{a}}=9$ and $\frac{\mathrm{a}}{\mathrm{e}}= \pm \frac{4}{\sqrt{3}}$

equation of tangent $y-\sqrt{3} x+\sqrt{3}=0$

by equation of tangent

Let slope $=\mathrm{S}=\sqrt{3}$

Constant $=-\sqrt{3}$

By condition of tangency

$ \Rightarrow 6=6 \mathrm{a}^2-9 \mathrm{a} $

$ \Rightarrow \mathrm{a}=2, \mathrm{~b}^2=9$

Equation of Hyperbola is

$\frac{x^2}{4}-\frac{y^2}{9}=1$ and for tangent

Point of contact is $(4,3 \sqrt{3})=\left(\mathrm{x}_0, \mathrm{y}_0\right)$

Now $\mathrm{e}=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$

Again product of focal distances

$\mathrm{m}=\left(\mathrm{x}_0 \mathrm{e}\right. $$ +\mathrm{a})\left(\mathrm{x}_0 \mathrm{e}-\mathrm{a}\right) $

$\mathrm{m}+4 \mathrm{e}^2 $$ =20 \mathrm{e}^2-\mathrm{a}^2 $

$ =20 \times \frac{13}{4}-4=61$

(There is a printing mistake in the equation of directrix $x= \pm \frac{4}{\sqrt{3}}$.

Corrected equation is $\mathrm{x}= \pm \frac{4}{\sqrt{13}}$ for directrix, as eccentricity must be greater than one, so question must be bonus)