$$y=m x \pm \sqrt{a^2 m^2-b^2}$$ is the tangent

$$\begin{aligned} & \Rightarrow m=\sqrt{3} \Rightarrow a^2 m^2-b^2=3 \\ & \Rightarrow 3 a^2-b^2=3 \end{aligned}$$

$$\begin{aligned} & \frac{2 b^2}{a}=9 \Rightarrow b^2=\frac{9 a}{2} \Rightarrow 3 a^2-\frac{9 a}{2}=3 \\ & \Rightarrow a^2-\frac{3}{2} a-1=0 \\ & \Rightarrow a=2 \text { or }-0.5 \text { (ignore) } \\ & \Rightarrow b=3 \\ & \Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1 \\ & \Rightarrow \text { Solving hyperbola and tangent } y=\sqrt{3 x}-\sqrt{3} \\ & x_0=4, y_0=3 \sqrt{3}, e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \\ & P F_1 \cdot P F_2= \\ & \sqrt{(4-\sqrt{13})^2+(3 \sqrt{3})^2} \sqrt{(4+\sqrt{13})^2+(3 \sqrt{3})^2} \\ & =\sqrt{(56-8 \sqrt{13})(56+8 \sqrt{13})} \\ & =\sqrt{2304}=48=m \\ & \Rightarrow 4 e^2+m=13+48=61 \end{aligned}$$