Question:
Let $O$ be the center and let $F$ be one of the foci of the ellipse $25x^2 +16 y^2 = 400$. A second ellipse, lying inside and tangent to the first ellipse, has its foci at $O$ and $F$. What is the length of the minor axis of this second ellipse?

Answer:
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 $(0, \pm 3).$

[asy]
unitsize(0.5 cm);

pair O = (0,0), F = (0,3);
path ellone = yscale(5)*xscale(4)*Circle((0,0),1);
path elltwo = shift((0,3/2))*yscale(7/2)*xscale(sqrt(10))*Circle((0,0),1);

draw((-5,0)--(5,0));
draw((0,-6)--(0,6));
draw(ellone);
draw(elltwo);

dot("$F$", F, E);
dot("$O$", O, NE);
dot("$(0,5)$", (0,5), NE);
[/asy]

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 \[\sqrt{7^2-3^2} = \boxed{2\sqrt{10}}.\]