Final answer:
The center of the ellipse is (2,2), the vertices are (4,2) and (0,2), and there are no real foci.
Step-by-step explanation:
The given equation is in the form of an ellipse. To identify the center, vertices, and foci of the ellipse, we need to compare the given equation with the standard form of an ellipse equation. The standard form is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center of the ellipse, a is the distance from the center to the vertices, and b is the distance from the center to the co-vertices. Comparing the given equation with the standard form, we can identify that the center of the ellipse is (2,2). The value under (x-2) determines the value of a which is 2, and the value under (y-2) determines the value of b which is 5. Therefore, the vertices of the ellipse are located at (2±a,2), which gives us (2±2,2) or (4,2) and (0,2). The foci of the ellipse are located at (2±c,2), where c is given by the equation c^2 = a^2 - b^2. Plugging in the values of a and b, we get c^2 = 4 - 25, which gives us c^2 = -21. Since c^2 is negative, the ellipse does not have any real foci.