Given equation of ellipse is $36x^2 + 4y^2$ = 144 i.e., $\frac{36x^2}{144} + \frac{4y^2}{144}$ = 1 ⇒ $\frac{x^2}{4} + \frac{y^2}{36}$ = 1 Clearly, 36 > 4 The equation of ellipse in standard form is $\frac{y^2}{a^2} + \frac{x^2}{b^2}$ = 1 ∴ $a^2$ = 36 ⇒ a = 6 and $b^2$ = 4 ⇒ b = 2 We know that c = $\sqrt{a^2-b^2}$ ⇒ c = $\sqrt{36-4}$ = $\sqrt{32}$ = $4\sqrt{2}$ ∴ Coordinates of foci are (0, ± c) i.e. (0, ± 4 $\sqrt{2}$) Coordinates of vertices are (0, ± a) i.e. (0, ± 6). Length of major axis = 2a = 2 × 6 = 12 Length of minor axis = 2b = 2 × 2 = 4 Eccentricity (e) = $\frac{c}{a}$ = $\frac{4\sqrt{2}}{6}$ = $\frac{2\sqrt{2}}{3}$ Length of latus rectum = $\frac{2b^2}{a}$ = $\frac{2\times 4}{6}$ = $\frac{4}{3}$