1. **Problem statement:** Find the length of the major and minor axes, coordinates of foci and vertices, and eccentricity for the ellipses: (i) $16x^2 + 25y^2 = 400$ (ii) $3x^2 + 2y^2 = 6$ 2. **General form and formulas:** The standard form of an ellipse centered at the origin is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $a > b > 0$. - Length of major axis = $2a$ - Length of minor axis = $2b$ - Coordinates of vertices: $(\pm a, 0)$ if major axis along x-axis, or $(0, \pm a)$ if along y-axis - Coordinates of foci: $(\pm c, 0)$ or $(0, \pm c)$ where $c = \sqrt{a^2 - b^2}$ - Eccentricity $e = \frac{c}{a}$ 3. **Ellipse (i):** $16x^2 + 25y^2 = 400$ Divide both sides by 400: $$\frac{16x^2}{400} + \frac{25y^2}{400} = 1 \Rightarrow \frac{x^2}{25} + \frac{y^2}{16} = 1$$ Here, $a^2 = 25$, $b^2 = 16$, so $a = 5$, $b = 4$. - Major axis length = $2a = 10$ - Minor axis length = $2b = 8$ - Since $a^2 > b^2$, major axis is along x-axis. - Vertices: $(\pm 5, 0)$ - Calculate $c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ - Foci: $(\pm 3, 0)$ - Eccentricity: $e = \frac{c}{a} = \frac{3}{5} = 0.6$ 4. **Ellipse (ii):** $3x^2 + 2y^2 = 6$ Divide both sides by 6: $$\frac{3x^2}{6} + \frac{2y^2}{6} = 1 \Rightarrow \frac{x^2}{2} + \frac{y^2}{3} = 1$$ Here, $a^2 = 3$, $b^2 = 2$, so $a = \sqrt{3} \approx 1.732$, $b = \sqrt{2} \approx 1.414$. - Major axis length = $2a = 2\sqrt{3} \approx 3.464$ - Minor axis length = $2b = 2\sqrt{2} \approx 2.828$ - Since $a^2 > b^2$, major axis is along y-axis. - Vertices: $(0, \pm \sqrt{3})$ - Calculate $c = \sqrt{a^2 - b^2} = \sqrt{3 - 2} = 1$ - Foci: $(0, \pm 1)$ - Eccentricity: $e = \frac{c}{a} = \frac{1}{\sqrt{3}} \approx 0.577$ **Final answers:** (i) Major axis length = 10, Minor axis length = 8, Vertices at $(\pm 5, 0)$, Foci at $(\pm 3, 0)$, Eccentricity = 0.6 (ii) Major axis length = $2\sqrt{3}$, Minor axis length = $2\sqrt{2}$, Vertices at $(0, \pm \sqrt{3})$, Foci at $(0, \pm 1)$, Eccentricity = $\frac{1}{\sqrt{3}}$