(2) If the foci and vertices of an ellipse are the vertices and foci of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{16}=1$, then the equation of the ellipse is $\qquad$
(2) $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ Hint: The two vertices of the hyperbola are $(0, \pm 3)$, and the two foci are $(0, \pm 5)$. Therefore, from the conditions, the two foci of the ellipse are $(0, \pm 3)$, and the two vertices are $(0, \pm 5)$. Thus, $c=3, a=5, b^{2}=a^{2}-c^{2}=16$, so the equation of the ellipse is $\frac{x^{2}}{16}+$ $\frac{y^{2}}{25}=1$