10. [-/6 Points] DETAILS MY NOTES LARPCALC10 10.3.036. Find the center, vertices, foci, and eccentricity of the ellipse. \[ \frac{(x+5)^{2}}{11}+\frac{(y-2)^{2}}{36}=1 \] center \( (x, y)=(\square) \) vertices \( (x, y)=(\square) \) (smaller \( y \)-value) \[ (x, y)=(\square) \text { (larger } y \text {-value) } \] foci \( (x, y)=(\square) \) (smaller \( y \)-value) \[ (x, y)=(\square) \text { (larger } y \text {-value) } \] eccentricity \( \square \)

The center of the ellipse can be found directly from the equation, which is given in the form \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\). Here, \(h = -5\) and \(k = 2\); therefore, the center is at \((-5, 2)\). The vertices, which are found by moving \(b\) units vertically from the center, are at \((-5, 2-6)\) and \((-5, 2+6)\), or \((-5, -4)\) and \((-5, 8)\). The foci are located along the major axis (the \(y\)-axis in this case) and can be found using the formula \(c = \sqrt{b^2 - a^2}\). Here, \(b = 6\) and \(a = \sqrt{11}\), so \(c = \sqrt{36 - 11} = \sqrt{25} = 5\). Thus, the foci are at \((-5, 2-5)\) and \((-5, 2+5)\), resulting in \((-5, -3)\) and \((-5, 7)\). The eccentricity \(e\) is calculated using the formula \(e = \frac{c}{b}\), yielding \(e = \frac{5}{6}\). Center: \( (x, y) = (-5, 2) \) Vertices: \( (x, y) = (-5, -4) \) (smaller \(y\)-value), \((-5, 8)\) (larger \(y\)-value) Foci: \( (x, y) = (-5, -3) \) (smaller \(y\)-value), \((-5, 7)\) (larger \(y\)-value) Eccentricity: \( \frac{5}{6} \)