Find the equation of an ellipse that satisfies the following conditions: Vertices $\;(\pm 13,0)$, Foci $\;(\pm 5,0)$

Question

$\begin{array}{1 1}\frac{x^2}{144}+\frac{y^2}{169}=1 \\ \frac{x^2}{196}+\frac{y^2}{144}=1 \\ \frac{x^2}{144}+\frac{y^2}{225}=1 \\ \frac{x^2}{225}+\frac{y^2}{169}=1\end{array} $