Since the foci (±5, 0) lie on x-axis. ∴ The equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 Now, length of major axis 2a = 26 ⇒ a = 13, Foci are (±5, 0) ⇒ c = 5 Also, we know that $c^2$ = $a^2 - b^2$ ⇒ 52 = $(13)^2 - b^2$ ⇒ $b^2$ = 169 – 25 = 144. Hence the required equation of ellipse is $\frac{x^2}{169}+\frac{y^2}{144}$ = 1