Clearly, the foci (0, ±5) lie on y-axis. ∴ The equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 Now, vertices are (0, ±13) ⇒ a = 13, foci are (0, ±5) ⇒ c = 5 Also, we know that $c^2$ = $a^2 - b^2$ ∴ $(5)^2$ = $(13)^2 - b^2$ ⇒ $b^2$ = 169 – 25 = 144. Hence the required equation of ellipse is $\frac{x^2}{144}+\frac{y^2}{169}$ = 1