Identify the type of the conic section represented by the equation \(x^2 - 4y^2 - 2x + 8y - 1 = 0\).

Step 1 ：Step 1: Rewrite the equation in standard form. The standard forms for the four conic sections are as follows: \[ \begin{align*} \text{Circle: } & (x - h)^2 + (y - k)^2 = r^2, \\ \text{Ellipse: } & \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, \\ \text{Hyperbola: } & \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1, \\ \text{Parabola: } & 4p(y - k) = (x - h)^2. \end{align*} \]

Step 2 ：Step 2: Rearrange the given equation to resemble one of the standard forms. \[ \begin{align*} x^2 - 4y^2 - 2x + 8y - 1 & = 0 \\ (x^2 - 2x) - 4(y^2 - 2y) & = 1 \\ (x - 1)^2 - 4(y - 1)^2 & = 1. \end{align*} \]

Step 3 ：Step 3: Compare the rearranged equation with the standard forms. The equation is in the form \(a(x - h)^2 - b(y - k)^2 = 1\), which is the standard form of a hyperbola.