Chapter 9 Quadratic Relations and Conic Sections - 9.5 Graph and Write Equations of Hyperbolas - 9.5 Exercises - Quiz for Lessons 9.4-9.5 - Page 648: 4

$\frac{x^2}{16}+\frac{y^2}{25}=1$

If the vertex of an ellipse is at $(0,\pm a)$ and the co-vertex is at $(\pm b,0)$, then the equation of the ellipse is: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ Hence here $a=5,b=-4$, thus the equaition is: $\frac{x^2}{16}+\frac{y^2}{25}=1$