The equation of the ellipse that has a center at $(10,3)$, a focus at $(13,3)$, and a vertex at $(15,3)$, is \[ \begin{array}{l} \frac{(x-C)^{2}}{A^{2}}+\frac{(y-D)^{2}}{B^{2}}=1 \\ \text { where } \\ A= \\ B= \\ C= \\ D= \end{array} \]

Step 1 ：The general equation of an ellipse is given by \(\frac{(x-C)^{2}}{A^{2}}+\frac{(y-D)^{2}}{B^{2}}=1\). The center of the ellipse is at point \((C, D)\), the distance from the center to the focus is \(c\), and the distance from the center to the vertex is \(a\).

Step 2 ：Given that the center is at \((10,3)\), we have \(C=10\) and \(D=3\).

Step 3 ：The focus is at \((13,3)\), so the distance from the center to the focus, \(c\), is \(13-10=3\).

Step 4 ：The vertex is at \((15,3)\), so the distance from the center to the vertex, \(a\), is \(15-10=5\).

Step 5 ：Since the focus and the vertex lie on the x-axis, the major axis of the ellipse is along the x-axis. Therefore, \(A=a=5\) and \(B\) can be found using the relationship \(a^2 = b^2 + c^2\) for an ellipse, where \(b\) is the distance from the center to the co-vertex.

Step 6 ：We can solve for \(B\) using the given values of \(a\) and \(c\).

Step 7 ：Given \(a = 5\) and \(c = 3\), we find that \(b = 4.0\).

Step 8 ：Final Answer: The equation of the ellipse is \(\boxed{\frac{(x-10)^{2}}{5^{2}}+\frac{(y-3)^{2}}{4^{2}}=1}\). So, \(A=5\), \(B=4\), \(C=10\), and \(D=3\).