jmc

The sum of the distances from $(14,-3)$ to the two foci is $$\sqrt{(14-2{)}^2+(-3-2{)}^2}+\sqrt{(14-2{)}^2+(-3-6{)}^2}=13+15=28.$$Therefore, the major axis has length $28.$ Since the distance between the foci is $\sqrt{(2-2{)}^2+(2-6{)}^2}=4,$ it follows that the length of the minor axis is $\sqrt{28^2-4^2}=4\sqrt{7^2-1}=4\sqrt{48}=16\sqrt{3}.$

The center of the ellipse is the midpoint of the segment between the foci, which is $(2,4).$ Since the foci and the center have the same $x$-coordinate, the major axis is parallel to the $y$-axis, and the minor axis is parallel to the $x$-axis. Putting all this together, we get the equation of the ellipse: $$\frac{(x-2{)}^2}{(8\sqrt{3}{)}^2}+\frac{(y-4{)}^2}{14^2}=1.$$Thus, $(a,b,h,k)=\boxed{(8\sqrt{3},14,2,4)}.$