Solve the equation $4x^2 - 4y^2 - 2x + 2y + 1 = 0$.

1. Problem Description

Solve the equation $4x^2 - 4y^2 - 2x + 2y + 1 = 0$.

2. Solution Steps

First, we rewrite the given equation as:

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

We can complete the square for the $x$ terms and the $y$ 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 $(\frac{1}{4})^2 = \frac{1}{16}$ inside the parenthesis.

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

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

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

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

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

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

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

Let $Y = y - \frac{1}{4}$ and $X = x - \frac{1}{4}$.

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

$4Y^2 - 4X^2 = 1$

This is a hyperbola.

If we rewrite this in terms of x and y, we have:

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

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

3. Final Answer

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

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

This is a hyperbola.