The problem is to analyze the equation $4x^2 - 4y^2 - 2x + 2y + 1 = 0$. We need to determine the type of conic section represented by this equation.

1. Problem Description

The problem is to analyze the equation $4x^2 - 4y^2 - 2x + 2y + 1 = 0$. We need to determine the type of conic section represented by this equation.

2. Solution Steps

First, rewrite the given equation by grouping the $x$ and $y$ terms:

$4x^2 - 2x - 4y^2 + 2y + 1 = 0$

Complete the square for the $x$ terms:

$4(x^2 - \frac{1}{2}x) - 4(y^2 - \frac{1}{2}y) + 1 = 0$

$4(x^2 - \frac{1}{2}x + \frac{1}{16}) - 4(\frac{1}{16}) - 4(y^2 - \frac{1}{2}y + \frac{1}{16}) + 4(\frac{1}{16}) + 1 = 0$

$4(x - \frac{1}{4})^2 - \frac{1}{4} - 4(y - \frac{1}{4})^2 + \frac{1}{4} + 1 = 0$

$4(x - \frac{1}{4})^2 - 4(y - \frac{1}{4})^2 + 1 = 0$

$4(x - \frac{1}{4})^2 - 4(y - \frac{1}{4})^2 = -1$

$4(y - \frac{1}{4})^2 - 4(x - \frac{1}{4})^2 = 1$

$(y - \frac{1}{4})^2 - (x - \frac{1}{4})^2 = \frac{1}{4}$

Let $X = x - \frac{1}{4}$ and $Y = y - \frac{1}{4}$. Then the equation becomes:

$Y^2 - X^2 = \frac{1}{4}$

$\frac{Y^2}{\frac{1}{4}} - \frac{X^2}{\frac{1}{4}} = 1$

This is a hyperbola centered at $(\frac{1}{4}, \frac{1}{4})$. The equation is of the form $\frac{Y^2}{a^2} - \frac{X^2}{b^2} = 1$, where $a^2 = \frac{1}{4}$ and $b^2 = \frac{1}{4}$. So $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

3. Final Answer

The equation represents a hyperbola.