Given equation of ellipse is $\frac{x^2}{49} + \frac{y^2}{36}$ = 1 Clearly, 49 > 36 The equation of ellipse in standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1 ∴ $a^2$ = 49 ⇒ a = 7 and $b^2$ = 36 ⇒ b = 6 We know that c = $\sqrt{a^2 - c^2}$ , c = $\sqrt{49 - 36}$ = $\sqrt{13}$ ∴ Coordinates of foci are (±c, 0) i.e. (± $\sqrt{13}$, 0) Coordinates of vertices are (±a, 0) i.e. (±7, 0). Length of major axis = 2a = 2 × 7 = 14 Length of minor axis = 2b = 2 × 6 = 12 Eccentricity (e) = $\frac{c}{a}$ = $\frac{\sqrt{13}}{7}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 36}{7}$ = $\frac{72}{7}$