jmc

Dividing by $400,$ we get the standard form of the equation for the first ellipse: $$\frac{x^2}{16}+\frac{y^2}{25}=1.$$Therefore, the semiaxes have lengths $\sqrt{16}=4$ and $\sqrt{25}=5,$ which means that the distance from the center $O=(0,0)$ to each focus is $\sqrt{5^2-4^2}=3.$ Since the vertical axis is longer than the horizontal axis, it follows that the foci of the first ellipse are at $(0,\pm 3).$



Without loss of generality, assume that $F=(0,3).$ Then the second ellipse must be tangent to the first ellipse at the point $(0,5).$ The sum of the distances from $(0,5)$ to the foci of the second ellipse is $2+5=7,$ so the length of the major axis of the second ellipse is $7.$ Since the distance between the foci of the second ellipse is $3,$ the length of the minor axis of the second ellipse is $$\sqrt{7^2-3^2}=\boxed{2\sqrt{10}}.$$