Given equation of ellipse is $16x^2 + y^2$ = 16 i.e., $\frac{16x^2}{16} + \frac{y^2}{16}$ = 1 ⇒ $\frac{x^2}{1} + \frac{y^2}{16}$ = 1 Clearly, 16 > 1 The equation of ellipse in standard form is $\frac{y^2}{a^2} + \frac{x^2}{b^2}$ = 1 ∴ $a^2$ = 16 ⇒ a = 4 and $b^2$ = 1 ⇒ b = 1 We know that c = $\sqrt{a^2 - b^2}$ ⇒ c = $\sqrt{16 - 1}$ = $\sqrt{15}$ ∴ Coordinates of foci are (0, ±c) i.e. (0, ± $\sqrt{15}$) Coordinates of vertices are (0, ± a) i.e. (0, ± 4). Length of major axis = 2a = 2 × 4 = 8 Length of minor axis = 2b = 2 × 1 = 2 Eccentricity (e) = $\frac{c}{a}$ = $\frac{\sqrt{15}}{4}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 1}{4}$ = $\frac{1}{2}$