jmc

Setting $x=\frac{y^2}{4},$ we get $$y=\frac{m{y}^2}{4}+b,$$or $m{y}^2-4y+4b=0.$ Since we have tangent, this quadratic has a double root, meaning that its discriminant is equal to 0. Thus, $$16-4(m)(4b)=0,$$or $bm=1.$

Setting $y=-\frac{x^2}{32},$ we get $$-\frac{x^2}{32}=mx+b,$$or $x^2+32mx+32b=0.$ Again, this quadratic has a double root. Thus, $$(32m{)}^2-4(32b)=0,$$or $b=8m^2.$

Substituting into $bm=1,$ we get $8m^3=1,$ so $m=\frac{1}{2}.$ Then $b=2,$ so the equation of the line is $\boxed{y=\frac{1}{2}x+2}.$