Hyperbola

Given $\; \frac{2 b^2}{a} = 9$ and $\; \frac{a}{e} = \pm \; \frac{4}{\sqrt{3}}$equation of tangent $y - \sqrt{3} x \mid + \sqrt{3} = 0$by equation of tangentLet slope $= S = \sqrt{3}$Constant $= -\sqrt{3}$By condition of tangency$\; \Rightarrow 6 = 6 a^2 - 9 a$$\; \Rightarrow a = 2, b^2 = 9$Equation of Hyperbola is$\; \frac{x^2}{4} - \; \frac{y^2}{9} = 1$ and for tangentPoint of contact is $(4, 3\sqrt{3}) = (x_0, y_0)$ Now e $= \sqrt{1 + \; \frac{9}{4}} = \; \frac{\sqrt{13}}{2}$Again product of focal distances$\; m = (x_0 e + a)(x_0 e - a)$$\; m + 4 e^2 = 20 e^2 - 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 $x = \pm \; \frac{4}{\sqrt{13}}$ for directrix, as eccentricity must be greater than one, so question must be bonus)