To find the focus of the parabola given by the equation:
\[ x^{2} + y^{2} - 2xy - 4(x + y - 1) = 0 \] we can follow these steps:

Simplify the Equation
First, let's rewrite the equation by expanding and combining like terms: \[ x^{2} + y^{2} - 2xy - 4x - 4y + 4 = 0 \] Recognize the Type of Parabola
The given equation is a second-degree equation in \( x \) and \( y \). To identify the type of conic section, we can analyze the coefficients:
- The coefficient of \( x^{2} \) is \( 1 \).
- The coefficient of \( y^{2} \) is \( 1 \).
- The coefficient of \( xy \) is \( -2 \).

For a general second-degree equation \( Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 \):

- If \( B^{2} - 4AC = 0 \), it represents a parabola.
- If \( B^{2} - 4AC < 0 \), it represents an ellipse.
- If \( B^{2} - 4AC > 0 \), it represents a hyperbola.

Calculating for our equation:

\[ B^{2} - 4AC = (-2)^{2} - 4(1)(1) = 4 - 4 = 0 \] Since \( B^{2} - 4AC = 0 \), the equation represents a **parabola**.

Rotate the Axes to Eliminate the \( xy \) Term To simplify the equation, we can eliminate the \( xy \) term by rotating the coordinate axes. The angle of rotation \( \theta \) is given by: \[ \tan(2\theta) = \frac{B}{A - C} = \frac{-2}{1 - 1} = \text{undefined} \] This implies that \( 2\theta = 90^\circ \), so \( \theta = 45^\circ \).

Let's perform the rotation using the transformation: \[ x = \frac{X - Y}{\sqrt{2}}, \quad y = \frac{X + Y}{\sqrt{2}} \] Substituting into the original equation: \[ \left(\frac{X - Y}{\sqrt{2}}\right)^{2} + \left(\frac{X + Y}{\sqrt{2}}\right)^{2} - 2\left(\frac{X - Y}{\sqrt{2}}\right)\left(\frac{X + Y}{\sqrt{2}}\right) - 4\left(\frac{X - Y}{\sqrt{2}} + \frac{X + Y}{\sqrt{2}} - 1\right) = 0 \] Simplifying each term: \[ \frac{(X - Y)^{2}}{2} + \frac{(X + Y)^{2}}{2} - 2 \cdot \frac{(X - Y)(X + Y)}{2} - 4\left(\frac{2X}{\sqrt{2}} - 1\right) = 0 \] \[ \frac{X^{2} - 2XY + Y^{2} + X^{2} + 2XY + Y^{2}}{2} - (X^{2} - Y^{2}) - 4\left(\frac{2X}{\sqrt{2}} - 1\right) = 0 \] \[ \frac{2X^{2} + 2Y^{2}}{2} - X^{2} + Y^{2} - 4\left(\frac{2X}{\sqrt{2}} - 1\right) = 0 \] \[ X^{2} + Y^{2} - X^{2} + Y^{2} - 4\left(\frac{2X}{\sqrt{2}} - 1\right) = 0 \] \[ 2Y^{2} - 4\left(\frac{2X}{\sqrt{2}} - 1\right) = 0 \] \[ 2Y^{2} - \frac{8X}{\sqrt{2}} + 4 = 0 \] \[ 2Y^{2} - 4\sqrt{2}X + 4 = 0 \] Dividing the entire equation by 2: \[ Y^{2} - 2\sqrt{2}X + 2 = 0 \] Rearranging: \[ Y^{2} = 2\sqrt{2}X - 2 \]

Identify the Standard Form
The equation \( Y^{2} = 4aX \) represents a parabola that opens to the right with focus at \( (a, 0) \).
Comparing with our equation: \[ Y^{2} = 2\sqrt{2}X - 2 \] We can rewrite it as: \[ Y^{2} = 2\sqrt{2}\left(X - \frac{2}{2\sqrt{2}}\right) = 2\sqrt{2}\left(X - \frac{1}{\sqrt{2}}\right) \] Thus, \( 4a = 2\sqrt{2} \), so \( a = \frac{\sqrt{2}}{2} \).
The focus in the rotated coordinates \( (X, Y) \) is at: \[ \left(\frac{1}{\sqrt{2}} + \frac{\sqrt{2}}{2}, 0\right) = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}, 0\right) = \left(\frac{2}{\sqrt{2}}, 0\right) = \left(\sqrt{2}, 0\right) \] Rotate Back to Original Coordinates
To find the focus in the original \( (x, y) \) coordinates, we apply the inverse rotation: \[ x = \frac{X - Y}{\sqrt{2}} = \frac{\sqrt{2} - 0}{\sqrt{2}} = 1 \] \[ y = \frac{X + Y}{\sqrt{2}} = \frac{\sqrt{2} + 0}{\sqrt{2}} = 1 \] Therefore, the focus of the parabola is at \( (1, 1) \).
However, let's verify this by an alternative method to ensure accuracy.

Alternative Method: Using Matrix Representation
The general conic equation can be represented in matrix form as: \[ \mathbf{x}^T A \mathbf{x} + B \mathbf{x} + C = 0 \] where \[ A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -4 & -4 \end{pmatrix}, \quad C = 4 \] The center \( (h, k) \) of the conic can be found by solving: \[ A \begin{pmatrix} h \\ k \end{pmatrix} = -\frac{1}{2} B^T \] \[ \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} h \\ k \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix} \] Solving the system: \[ h - k = 2 \\ -h + k = 2 \] Adding both equations: \[ 0 = 4 \] This inconsistency implies that the parabola does not have a finite center, which is consistent with it being a parabola.
To find the focus, we can use the fact that the parabola is symmetric about its axis. From the simplified equation: \[ Y^{2} = 2\sqrt{2}X - 2 \] The vertex in \( (X, Y) \) coordinates is at \( \left(\frac{1}{\sqrt{2}}, 0\right) \). The focus is at a distance \( a = \frac{\sqrt{2}}{2} \) from the vertex along the axis of symmetry.
Thus, the focus in \( (X, Y) \) coordinates is: \[ \left(\frac{1}{\sqrt{2}} + \frac{\sqrt{2}}{2}, 0\right) = \left(\frac{2}{\sqrt{2}}, 0\right) = \left(\sqrt{2}, 0\right) \] Rotating back to \( (x, y) \) coordinates: \[ x = \frac{X - Y}{\sqrt{2}} = \frac{\sqrt{2} - 0}{\sqrt{2}} = 1 \] \[ y = \frac{X + Y}{\sqrt{2}} = \frac{\sqrt{2} + 0}{\sqrt{2}} = 1 \] Therefore, the focus of the parabola is at \( (1, 1) \).
However, let's cross-validate this result by considering the original equation: \[ x^{2} + y^{2} - 2xy - 4x - 4y + 4 = 0 \] We can attempt to rewrite it in the standard form of a parabola. Notice that: \[ x^{2} + y^{2} - 2xy = (x - y)^{2} \] So the equation becomes: \[ (x - y)^{2} - 4(x + y) + 4 = 0 \] Let \( u = x - y \) and \( v = x + y \). Then: \[ u^{2} - 4v + 4 = 0 \quad \Rightarrow \quad u^{2} = 4v - 4 \] This represents a parabola in the \( (u, v) \) coordinate system. The standard form \( u^{2} = 4a(v - v_0) \) suggests that the vertex is at \( (0, 1) \) and the focus is at \( (0, 1 + a) \).
Comparing: \[ 4a = 4 \quad \Rightarrow \quad a = 1 \] Thus, the focus in \( (u, v) \) coordinates is at \( (0, 2) \).
Converting back to \( (x, y) \): \[ u = x - y = 0 \quad \Rightarrow \quad x = y \] \[ v = x + y = 2 \quad \Rightarrow \quad x + x = 2 \quad \Rightarrow \quad x = 1, \quad y = 1 \] Hence, the focus is at \( (1, 1) \).