$\begin{array}{l} 3{x^2} + 9{y^2} = 36 \Leftrightarrow \dfrac{{{x^2}}}{{12}} + \dfrac{{{y^2}}}{4} = 1\\ \to {a^2} = 12,{b^2} = 4 \to a = 2\sqrt 3 ,b = 2\left( {a > b > 0} \right)\\ \to {c^2} = {a^2} - {b^2} = 8 \Rightarrow c = 2\sqrt 2 \\ \to {F_1}{F_2} = 2c = 4\sqrt 2 \\ \to {F_1}\left( { - 2\sqrt 2 ;0} \right),{F_2}\left( {2\sqrt 2 ;0} \right) \end{array}$