We are given the equation $4x^2 - 4y^2 - 2x + 2y + 1 = 0$ and we need to determine the type of conic section this equation represents and find its standard form.

1. Problem Description

We are given the equation $4x^2 - 4y^2 - 2x + 2y + 1 = 0$ and we need to determine the type of conic section this equation represents and find its standard form.

2. Solution Steps

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

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

Next, we complete the square for the $x$ and $y$ terms.

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

To complete the square for $x^2 - \frac{1}{2}x$, we need to add and subtract $(\frac{1}{2} \cdot \frac{1}{2})^2 = (\frac{1}{4})^2 = \frac{1}{16}$.

To complete the square for $y^2 - \frac{1}{2}y$, we need to add and subtract $(\frac{1}{2} \cdot \frac{1}{2})^2 = (\frac{1}{4})^2 = \frac{1}{16}$.

So,

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

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

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

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

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

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

Divide by -1:

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

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

Divide by 4:

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

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

This is a hyperbola centered at $(\frac{1}{4}, \frac{1}{4})$.

3. Final Answer

The equation represents a hyperbola with the standard form

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