The centre of the conic `14x^(2)-4xy+11y^(2)-44x-58y+71=0,` is

Answer

To find the center of the conic given by the equation \( 14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0 \), we will follow these steps:

### Step 1: Identify the coefficients
The general form of a conic section is given by:
\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
From the given equation, we can identify:
- \( A = 14 \)
- \( B = -4 \)
- \( C = 11 \)
- \( D = -44 \)
- \( E = -58 \)
- \( F = 71 \)

### Step 2: Find the center using derivatives
The center of the conic can be found by solving the following equations:
1. \(\frac{\partial f}{\partial x} = 0\)
2. \(\frac{\partial f}{\partial y} = 0\)

### Step 3: Differentiate with respect to \( x \)
Differentiating the equation with respect to \( x \):
\[ \frac{\partial f}{\partial x} = 2Ax + By + D = 0 \]
Substituting the values:
\[ 2(14)x + (-4)y - 44 = 0 \implies 28x - 4y - 44 = 0 \]
This simplifies to:
\[ 28x - 4y = 44 \quad \text{(Equation 1)} \]

### Step 4: Differentiate with respect to \( y \)
Differentiating the equation with respect to \( y \):
\[ \frac{\partial f}{\partial y} = Bx + 2Cy + E = 0 \]
Substituting the values:
\[ (-4)x + 2(11)y - 58 = 0 \implies -4x + 22y - 58 = 0 \]
This simplifies to:
\[ -4x + 22y = 58 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( 28x - 4y = 44 \)
2. \( -4x + 22y = 58 \)

We can solve these equations simultaneously. Let's multiply Equation 2 by 7 to align the coefficients of \( y \):
\[ -28x + 154y = 406 \]
Now we can add this to Equation 1:
\[ (28x - 4y) + (-28x + 154y) = 44 + 406 \]
This simplifies to:
\[ 150y = 450 \implies y = 3 \]

### Step 6: Substitute \( y \) back to find \( x \)
Now substitute \( y = 3 \) back into Equation 1:
\[ 28x - 4(3) = 44 \implies 28x - 12 = 44 \implies 28x = 56 \implies x = 2 \]

### Conclusion
Thus, the center of the conic is at the point \( (x, y) = (2, 3) \).