12. [-/6 Points] DETAILS MY NOTES LARPCALC10 10.3.040. Find the center, vertices, foci, and eccentricity of the ellipse. \[ \begin{array}{l} 9 x^{2}+4 y^{2}-90 x+16 y+97=0 \\ \text { center } \quad(x, y)=(\square) \\ \text { vertices } \quad(x, y)=(\square) \text { (smaller } y \text {-value) } \\ (x, y)=(\square) \text { (larger } y \text {-value) } \\ \text { foci } \quad(x, y)=(\square) \text { (smaller } y \text {-value) } \\ (x, y)=(\square) \text { (larger } y \text {-value) } \\ \text { eccentricity } \end{array} \]

To find the characteristics of the ellipse represented by the equation \( 9x^2 + 4y^2 - 90x + 16y + 97 = 0 \), we first reorganize it into the standard form. Completing the square for both \( x \) and \( y \), we get: 1. Rearranging and completing the square leads to the standard form \( \frac{(x - 5)^2}{25} + \frac{(y + 2)^2}{16} = 1 \). From this, we see that the center is at \( (5, -2) \). 2. The vertices are found using the lengths of the semi-major and semi-minor axes. For our ellipse, the vertices are at \( (5, 2) \) (larger y-value) and \( (5, -6) \) (smaller y-value). 3. The foci can be calculated with the formula \( c = \sqrt{a^2 - b^2} \), where \( c = \sqrt{25 - 16} = 3 \). Thus, the foci are at \( (5, 1) \) (larger y-value) and \( (5, -5) \) (smaller y-value). 4. The eccentricity \( e \) is given by \( e = \frac{c}{a} = \frac{3}{5} \). So, summarizing: - Center: \( (5, -2) \) - Vertices: \( (5, 2), (5, -6) \) - Foci: \( (5, 1), (5, -5) \) - Eccentricity: \( \frac{3}{5} \) Let’s break it down with a little fun: imagine your favorite cartoon character navigating through a cosmic elliptical pathway. Each hiccup and swoop to the vertices is mastered, while the foci whisper sweet secrets of gravity pulling them in! Isn’t math just like a thrilling adventure?