The co-ordinates of the centre of the hyperbola, `x^2+3x y+3y^2+2x+3y+2=0` is `(-1,0)` (b) (1, 0) `(-1,1)` (d) `(1,-1)`

Options

A. `(-1, 0)`
B. `(1, 0)`
C. `(-1, 1)`
D. `(1, -1)`

Answer

To find the coordinates of the center of the hyperbola given by the equation \( x^2 + 3xy + 3y^2 + 2x + 3y + 2 = 0 \), we can use partial differentiation. Here’s a step-by-step solution:

### Step 1: Write down the equation
The equation of the hyperbola is:
\[ x^2 + 3xy + 3y^2 + 2x + 3y + 2 = 0 \]

### Step 2: Partial differentiate with respect to \( x \)
We treat \( y \) as a constant and differentiate the equation with respect to \( x \):
\[ \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial x}(3xy) + \frac{\partial}{\partial x}(3y^2) + \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial x}(3y) + \frac{\partial}{\partial x}(2) = 0 \]
Calculating each term:
- \( \frac{\partial}{\partial x}(x^2) = 2x \)
- \( \frac{\partial}{\partial x}(3xy) = 3y \)
- \( \frac{\partial}{\partial x}(3y^2) = 0 \) (since \( y \) is constant)
- \( \frac{\partial}{\partial x}(2x) = 2 \)
- \( \frac{\partial}{\partial x}(3y) = 0 \)
- \( \frac{\partial}{\partial x}(2) = 0 \)

Thus, we have:
\[ 2x + 3y + 2 = 0 \quad \text{(Equation 1)} \]

### Step 3: Partial differentiate with respect to \( y \)
Now, we treat \( x \) as a constant and differentiate the equation with respect to \( y \):
\[ \frac{\partial}{\partial y}(x^2) + \frac{\partial}{\partial y}(3xy) + \frac{\partial}{\partial y}(3y^2) + \frac{\partial}{\partial y}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial y}(2) = 0 \]
Calculating each term:
- \( \frac{\partial}{\partial y}(x^2) = 0 \)
- \( \frac{\partial}{\partial y}(3xy) = 3x \)
- \( \frac{\partial}{\partial y}(3y^2) = 6y \)
- \( \frac{\partial}{\partial y}(2x) = 0 \)
- \( \frac{\partial}{\partial y}(3y) = 3 \)
- \( \frac{\partial}{\partial y}(2) = 0 \)

Thus, we have:
\[ 3x + 6y + 3 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( 2x + 3y + 2 = 0 \)
2. \( 3x + 6y + 3 = 0 \)

#### Step 4.1: Multiply Equation 1 by 2
\[ 4x + 6y + 4 = 0 \]

#### Step 4.2: Subtract Equation 2 from the modified Equation 1
\[ (4x + 6y + 4) - (3x + 6y + 3) = 0 \]
This simplifies to:
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

#### Step 4.3: Substitute \( x = -1 \) back into Equation 1
\[ 2(-1) + 3y + 2 = 0 \]
This simplifies to:
\[ -2 + 3y + 2 = 0 \quad \Rightarrow \quad 3y = 0 \quad \Rightarrow \quad y = 0 \]

### Final Answer
The coordinates of the center of the hyperbola are:
\[ (-1, 0) \]