We have been given an equation of ellipse $\frac{x^2}{36}+\frac{y^2}{16}$ = 1 Clearly, 36 > 16 The equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 ∴ $a^2$ = 36 ⇒ a = 6 and $b^2$ = 16 ⇒ b = 4 We know that c = $\sqrt{a^2-b^2}$ ⇒ c = $\sqrt{36-16}$ = $\sqrt{20}$ ∴ Coordinates of foci are (±c, 0) i.e. (± $\sqrt{20}$, 0) Coordinates of vertices are (± a, 0) i.e. (± 6, 0). Length of major axis = 2a = 2 × 6 = 12 Length of minor axis = 2b = 2 × 4 = 8 Eccentricity (e) = $\frac{c}{a}$ = $\frac{\sqrt{20}}{6}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 16}{6}$ = $\frac{16}{3}$