$\begin{array}{l} \left\{ \begin{array}{l} {a^2} - {b^2} = 3\\ \dfrac{1}{{{a^2}}} + \dfrac{3}{{4{b^2}}} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {a^2} = 3 + {b^2}(1)\\ \dfrac{1}{{3 + {b^2}}} + \dfrac{3}{{4{b^2}}} = 1(2) \end{array} \right.\\ \left( 1 \right) \to \left( 2 \right):4{b^2} + 9 + 3{b^2} = \left( {3 + {b^2}} \right)4{b^2}\\ \Leftrightarrow 4{b^4} + 5{b^2} - 9 = 0 \Leftrightarrow \left( {{b^2} - 1} \right)\left( {4{b^2} + 9} \right) = 0\\ \Leftrightarrow b = \pm 1 \Leftrightarrow {a^2} = 4 \Leftrightarrow a = \pm 2\\ \Rightarrow \left( {a;b} \right) = \left( {1;2} \right),\left( { - 1; - 2} \right),\left( {1; - 2} \right),\left( { - 1;2} \right) \end{array}$
