The focus of the conic `x^(2)-6x+4y+1=0` is

Options

A. (2,3)
B. (3,2)
C. (3,1)
D. (1,4)

Answer

To find the focus of the conic given by the equation \( x^2 - 6x + 4y + 1 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
First, we want to rearrange the equation into a standard form. We start with:
\[ x^2 - 6x + 4y + 1 = 0 \]
Rearranging gives:
\[ x^2 - 6x + 1 = -4y \]

### Step 2: Completing the Square
Next, we complete the square for the \( x \) terms. The expression \( x^2 - 6x \) can be rewritten as:
\[ x^2 - 6x = (x - 3)^2 - 9 \]
Substituting this back into the equation gives:
\[ (x - 3)^2 - 9 + 1 = -4y \]
Simplifying this, we have:
\[ (x - 3)^2 - 8 = -4y \]
or
\[ (x - 3)^2 = -4y + 8 \]
which can be rearranged to:
\[ (x - 3)^2 = -4(y - 2) \]

### Step 3: Identifying the Conic Type
The equation \( (x - 3)^2 = -4(y - 2) \) is in the standard form of a parabola that opens downwards:
\[ (x - h)^2 = 4p(y - k) \]
where \( (h, k) \) is the vertex and \( p \) is the distance from the vertex to the focus.

### Step 4: Finding the Vertex
From the equation, we can identify:
- \( h = 3 \)
- \( k = 2 \)
- \( 4p = -4 \) which gives \( p = -1 \)

Thus, the vertex of the parabola is at \( (3, 2) \).

### Step 5: Finding the Focus
The focus of a parabola is located at \( (h, k + p) \). Therefore:
\[ \text{Focus} = (3, 2 - 1) = (3, 1) \]

### Conclusion
The focus of the conic \( x^2 - 6x + 4y + 1 = 0 \) is:
\[ \boxed{(3, 1)} \]