[tex]\bf \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1
\qquad
\begin{array}{llll}
center\ ({{ h}},{{ k}})\\\\
vertices\ ({{ h}}, {{ k}}\pm a)\\\\
foci\ (h\ ,\ k\pm \sqrt{a^2-b^2})
\end{array} \\\\
-----------------------------\\\\
\cfrac{x^2}{16}+\cfrac{y^2}{25}=1\implies \cfrac{(x-0)^2}{4^2}+\cfrac{(y-0)^2}{5^2}=1[/tex]
in an ellipse, the "a" component, is always the bigger denominator, in this case the 5, which is under the "y", meaning the "y-axis" is the major axis, and thus the foci are [tex]c=\sqrt{a^2-b^2}[/tex] distance from "k" coordinate
