Graph the ellipse and give the location of its foci. \[ \frac{(x-2)^{2}}{49}+\frac{(y-8)^{2}}{9}=1 \]

We start with the ellipse equation \[ \frac{(x-2)^2}{49} + \frac{(y-8)^2}{9} = 1. \] The standard form of an ellipse is \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \] so we identify the center as \[ (h,k) = (2,8). \] Here, \[ a^2 = 49 \quad \text{and} \quad b^2 = 9, \] so \[ a = 7 \quad \text{and} \quad b = 3. \] Since \(a > b\) and the term with \(a^2\) is under \((x-2)^2\), the major axis is horizontal. The foci are located along the major axis at a distance \(c\) from the center, where \[ c^2 = a^2 - b^2. \] Thus, \[ c^2 = 49 - 9 = 40, \] \[ c = \sqrt{40} = 2\sqrt{10}. \] Hence, the foci have coordinates \[ (2 \pm 2\sqrt{10},\, 8). \] To graph the ellipse: 1. Plot the center at \( (2,8) \). 2. From the center, move 7 units left and right along the \(x\)-axis to plot the vertices at \( (2\pm 7,\,8) \) or \( (-5,8) \) and \( (9,8) \). 3. Move 3 units up and down from the center along the \(y\)-axis to plot the co-vertices at \( (2,8\pm3) \) or \( (2,5) \) and \( (2,11) \). 4. Sketch the ellipse with these four points and note that it is wider than it is tall. 5. Mark the foci at \( (2+2\sqrt{10},\,8) \) and \( (2-2\sqrt{10},\,8) \). Thus, the final answer for the foci is: \[ (2+2\sqrt{10},\,8) \quad \text{and} \quad (2-2\sqrt{10},\,8). \]