Find the equation of the ellipse whose vertices are $(±13,\,0)$ and foci are $(±5,\,0)$.

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since the vertices are on $x-$ axis, the equation will be of the form

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where a is the semi-major axis.

Given that $a=13$ , $c=\pm 5$

Therefore, from the relation $c^{2}=a^{2}-b^{2},$ we get

$25=169-b^{2}$,  i.e., $b=12$

Hence the equation of the ellipse is $\frac{x^{2}}{169}+\frac{y^{2}}{144}=1$

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