The focus of the parabola `y^(2)-x-2y+2=0` is

Options

A. `(1//4,0)`
B. `(1//2)`
C. `(3//4,1)`
D. `(5//4, 1)`

Explanation

To find the focus of the parabola given by the equation \( y^2 - x - 2y + 2 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
Start with the given equation:
\[ y^2 - x - 2y + 2 = 0 \]
Rearranging it gives:
\[ y^2 - 2y + 2 = x \]

### Step 2: Complete the square for the \( y \) terms
The expression \( y^2 - 2y \) can be completed to a square:
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Substituting this back into the equation gives:
\[ (y - 1)^2 - 1 + 2 = x \]
This simplifies to:
\[ (y - 1)^2 = x - 1 \]

### Step 3: Identify the standard form of the parabola
The equation \( (y - 1)^2 = x - 1 \) can be rewritten as:
\[ (y - 1)^2 = 1(x - 1) \]
This matches the standard form of a parabola \( (y - k)^2 = 4a(x - h) \), where:
- \( h = 1 \)
- \( k = 1 \)
- \( 4a = 1 \)

### Step 4: Determine the values of \( h \), \( k \), and \( a \)
From the equation, we have:
- \( h = 1 \)
- \( k = 1 \)
- \( 4a = 1 \) implies \( a = \frac{1}{4} \)

### Step 5: Find the coordinates of the focus
The coordinates of the focus of a parabola in this form are given by:
\[ (h + a, k) \]
Substituting the values we found:
\[ \text{Focus} = \left(1 + \frac{1}{4}, 1\right) = \left(\frac{5}{4}, 1\right) \]

### Final Answer
Thus, the focus of the parabola is:
\[ \left(\frac{5}{4}, 1\right) \]

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