jmc

The center of the hyperbola must lie at the point $(t,2),$ for some $t>4.$ Then the distance from the center to each vertex is $a=t-4,$ and the distance from the center to each focus is $c=t-3.$ Therefore, we have $$b=\sqrt{c^2-a^2}=\sqrt{(t-3{)}^2-(t-4{)}^2}=\sqrt{2t-7}.$$The equation for the hyperbola can be written in standard form as $$\frac{(x-t{)}^2}{a^2}-\frac{(y-2{)}^2}{b^2}=1.$$Then the equations of the asymptotes are $\frac{x-t}{a}=\pm \frac{y-2}{b},$ or $y=2\pm \frac{b}{a}(x-t).$ Thus, the slopes of the asymptotes are $\pm \frac{b}{a}.$ Since $a>0$ and $b>0,$ we must have $\frac{b}{a}=\frac{\sqrt{2}}{2},$ or $b\sqrt{2}=a.$ Thus, $$\sqrt{2t-7}\cdot \sqrt{2}=t-4.$$Squaring both sides of this equation gives $$2(2t-7)=(t-4{)}^2,$$or $t^2-12t+30=0.$ By the quadratic formula, $$t=\frac{12\pm \sqrt{12^2-4\cdot 30}}{2}=6\pm \sqrt{6}.$$Because $t>4$ and $6-\sqrt{6}<6-2=4,$ we must have $t=\boxed{6+\sqrt{6}}.$