Find the center, vertices, foci, and eccentricity of the ellipse. \[ \begin{array}{l} \frac{x^{2}}{9}+\frac{y^{2}}{100}=1 \\ \text { center } \quad(x, y)=(0,0,) \\ \text { vertices } \quad(x, y)=(\square) \text { (smaller } y \text {-value) } \\ (x, y)=(\square \times) \text { (larger } y \text {-value) } \\ \text { foci } \\ (x, y)=(\square \times) \text { (smaller } y \text {-value) } \\ (x, y)=(\square \times) \text { (larger } y \text {-value) } \\ \text { eccentricity } \end{array} \]

To analyze the given ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{100}=1\), we can start by identifying key characteristics. The center of the ellipse is at \((0,0)\), as noted. The vertices are determined by the values under the squares in the equation. The larger denominator indicates that the ellipse is stretched vertically. Hence, the vertices are located at \((0, 10)\) and \((0, -10)\). Now, for the foci, we need to compute \(c\), where \(c = \sqrt{b^2 - a^2}\) with \(a^2=100\) and \(b^2=9\). This gives us \(c = \sqrt{100 - 9} = \sqrt{91} \approx 9.54\). Therefore, the foci are at approximately \((0, 9.54)\) and \((0, -9.54)\). Finally, the eccentricity \(e\) can be computed using \(e = \frac{c}{a} = \frac{\sqrt{91}}{10} \approx 0.954\). So to summarize, we have: - \( (0,10) \) (larger y-value) and \( (0,-10) \) (smaller y-value) as vertices - \( (0, 9.54) \) (larger y-focus) and \( (0, -9.54) \) (smaller y-focus) - Eccentricity \( e \approx 0.954 \) Here is the formatted result for the blanks you provided: \[ \begin{array}{l} \text { center } \quad(x, y)=(0,0,) \\ \text { vertices } \quad(x, y)=(0, -10) \text { (smaller } y \text {-value) } \\ (x, y)=(0, 10) \text { (larger } y \text {-value) } \\ \text { foci } \\ (x, y)=(0, -9.54) \text { (smaller } y \text {-value) } \\ (x, y)=(0, 9.54) \text { (larger } y \text {-value) } \\ \text { eccentricity } \approx 0.954 \end{array} \]