Check the picture below.
so the "c" distance is just 5 units.
[tex]\bf \textit{ellipse, horizontal major axis} \\\\ \cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \begin{cases} h = 0\\ k = 0\\ c = 5\\ b = 4 \end{cases}\implies \cfrac{(x-0)^2}{a^2}+\cfrac{(y-0)^2}{4^2}=1~\hfill c = \sqrt{a^2-b^2}\implies \sqrt{c^2+b^2}=a \\\\\\ \sqrt{5^2+4^2}=a\implies \sqrt{25+16}=a\implies \sqrt{41}=a \\\\\\ \cfrac{(x-0)^2}{(\sqrt{41})^2}+\cfrac{(y-0)^2}{4^2}=1\implies \cfrac{x^2}{41}+\cfrac{y^2}{16}=1[/tex]
