jmc

Dividing the equation of the ellipse by $225,$ we get $$\frac{x^2}{9}+\frac{y^2}{25}=1.$$Therefore, the semi-major axis has length $\sqrt{25}=5$ and is vertical, while the semi-minor axis has length $\sqrt{9}=3$ and is horizontal. This means that the endpoints of the major axis are $(0,\pm 5).$ Also, the distance from each focus of the ellipse to the center (the origin) is $\sqrt{5^2-3^2}=4,$ so the foci of the ellipse are at $(0,\pm 4).$

Now, we know that the hyperbola has its vertices at $(0,\pm 4)$ and its foci at $(0,\pm 5).$ Since these points all lie along the $y-$axis, the equation of the hyperbola must take the form $$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$$(as opposed to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$). Since the vertices are at $(0,\pm 4),$ we have $a=4.$ The distance from each focus to the center of the hyperbola (the origin) is $c=5,$ so we have $b=\sqrt{c^2-a^2}=3.$ Therefore, the equation of the hyperbola is $$\frac{y^2}{16}-\frac{x^2}{9}=1,$$or $9y^2-16x^2=144.$ Now we want to solve the system $$\begin{array}{rl}25x^2+9y^2& =225,\\ 9y^2-16x^2& =144.\end{array}$$Subtracting these equations, we get $41x^2=81,$ so $x^2=\frac{81}{41}.$ That is, the coordinates $(s,t)$ of the intersection point satisfy $s^2=\boxed{\frac{81}{41}}.$