jmc

Because the hyperbola has center at $(0,0)$ and intersects the $x-$axis at $(-2,0)$, it must open horizontally, and $(-2,0)$ must be one of its vertices. Therefore, it has an equation of the form $$\frac{x^2}{2^2}-\frac{y^2}{b^2}=1$$for some $b>0.$ Setting $x=-3$ and $y=4,$ we get the equation $$\frac{9}{4}-\frac{16}{b^2}=1,$$which gives $b^2=\frac{64}{5}.$ Therefore, the equation of the hyperbola is $$\frac{x^2}{4}-\frac{5y^2}{64}=1.$$Setting $x=t$ and $y=2,$ we get $$\frac{t^2}{4}-\frac{5}{16}=1,$$which gives $t^2=\boxed{\frac{21}{4}}.$