jmc

Solving the system $y=2x+5$ and $y=-2x+1,$ we get $(x,y)=(-1,3).$ Therefore, the asymptotes of the hyperbola intersect at $(-1,3),$ which must be the center of the hyperbola. Therefore, $(h,k)=(-1,3),$ so the equation of the hyperbola is $$\frac{(y-3{)}^2}{a^2}-\frac{(x+1{)}^2}{b^2}=1$$for some $a$ and $b.$ The equations of the asymptotes are therefore $$\frac{y-3}{a}=\pm \frac{x+1}{b},$$or $$y=3\pm \frac{a}{b}(x+1).$$Therefore, the slopes of the asymptotes are $\pm \frac{a}{b}.$ Because $a$ and $b$ are positive, we must have $\frac{a}{b}=2,$ so $a=2b.$ Therefore, the equation of the hyperbola is $$\frac{(y-3{)}^2}{4b^2}-\frac{(x+1{)}^2}{b^2}=1.$$To find $b,$ we use the fact that the hyperbola passes through $(0,7).$ Setting $x=0$ and $y=7$ gives the equation $$\frac{(7-3{)}^2}{4b^2}-\frac{(0+1{)}^2}{b^2}=1,$$or $\frac{3}{b^2}=1.$ Thus, $b=\sqrt{3},$ and so $a=2b=2\sqrt{3}.$ Hence the equation of the hyperbola is $$\frac{(y-3{)}^2}{12}-\frac{(x+1{)}^2}{3}=1,$$and $a+h=\boxed{2\sqrt{3}-1}.$