Find the standard form of the equation of the hyperbola satisfying the given conditions. Center: $(1,-2)$; focus: $(10,-2)$; vertex: $(9,-2)$

Step 1 ：We are given that the center of the hyperbola is at (1, -2), one focus is at (10, -2), and one vertex is at (9, -2).

Step 2 ：The distance from the center to the focus is 9 units, and the distance from the center to the vertex is 8 units. Since the focus and vertex have the same y-coordinate as the center, the transverse axis is horizontal.

Step 3 ：The standard form of the equation of a hyperbola with a horizontal transverse axis is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), where (h, k) is the center, a is the distance from the center to a vertex, and b is the distance from the center to a co-vertex.

Step 4 ：We can find b by using the relationship \(c^2 = a^2 + b^2\), where c is the distance from the center to a focus. Solving for b gives \(b = \sqrt{c^2 - a^2}\).

Step 5 ：Substituting the given values, we find that \(b = \sqrt{9^2 - 8^2} = \sqrt{17}\), which is approximately 4.123105625617661.

Step 6 ：Substituting h = 1, k = -2, a = 8, and b = \sqrt{17} into the equation for a hyperbola with a horizontal transverse axis, we find the standard form of the equation of the hyperbola to be \(\frac{(x - 1)^2}{64} - \frac{(y + 2)^2}{17} = 1\).

Step 7 ：Final Answer: The standard form of the equation of the hyperbola is \(\boxed{\frac{(x - 1)^2}{64} - \frac{(y + 2)^2}{17} = 1}\)