Given equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{9}$ = 1 Clearly, 16 > 9 The equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 ∴ $a^2$ = 16 ⇒ a = 4 and $b^2$ = 9 ⇒ b = 3 We know that c = $\sqrt{a^2-b^2}$ ⇒ c = $\sqrt{16-9}$ = $\sqrt{7}$ ∴ Coordinates of foci are (±c, 0) i.e. (± $\sqrt{7}$ , 0) Coordinates of vertices are (±a, 0) i.e. (±4, 0). Length of major axis = 2a = 2 × 4 = 8 Length of minor axis = 2b = 2 × 3 = 6 Eccentricity (e) = $\frac{c}{a}$ = $\frac{\sqrt{7}}{4}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2\times 9}{4}$ = $\frac{9}{2}$