The equation of a hyperbola conjugate to the hyperbola `x^(2)+3xy+2y^(2)+2x+3y=0` is

Options

A. `x^(2)+3xy+2y^(2)+2x+3y+1=0`
B. `x^(2)+3xy+2y^(2)+2x+3y+2=0`
C. `x^(2)+3xy+2y^(2)+2x+3y+3=0`
D. `x^(2)+3xy+2y^(2)+2x+3y+4=0`

Answer

To find the equation of the hyperbola conjugate to the given hyperbola \( x^2 + 3xy + 2y^2 + 2x + 3y = 0 \), we can follow these steps:

### Step 1: Identify the Given Hyperbola Equation
The given hyperbola equation is:
\[ H: x^2 + 3xy + 2y^2 + 2x + 3y = 0 \]

### Step 2: Use the Property of Hyperbolas
The property states that:
\[ 2 \times \text{(asymptotic equation)} = \text{(hyperbola equation)} + \text{(conjugate hyperbola equation)} \]
We need to find the asymptotic equation first.

### Step 3: Write the Asymptotic Equation
Let’s express the asymptotic equation as:
\[ x^2 + 3xy + 2y^2 + 2x + 3y + c = 0 \]
where \( c \) is a constant we need to determine.

### Step 4: Partial Differentiation
To find the value of \( c \), we partially differentiate the hyperbola equation with respect to \( x \) and \( y \).

1. **Partial differentiation with respect to \( x \)**:
\[ \frac{\partial H}{\partial x} = 2x + 3y + 2 = 0 \]

2. **Partial differentiation with respect to \( y \)**:
\[ \frac{\partial H}{\partial y} = 3x + 4y + 3 = 0 \]

### Step 5: Solve the System of Equations
Now we have a system of equations:
1. \( 2x + 3y + 2 = 0 \)
2. \( 3x + 4y + 3 = 0 \)

Multiply the first equation by 3 and the second by 2 to eliminate \( x \):
1. \( 6x + 9y + 6 = 0 \)
2. \( 6x + 8y + 6 = 0 \)

Subtract the second from the first:
\[ (6x + 9y + 6) - (6x + 8y + 6) = 0 \implies y = 0 \]

### Step 6: Substitute \( y \) to Find \( x \)
Substituting \( y = 0 \) into the first equation:
\[ 2x + 3(0) + 2 = 0 \implies 2x + 2 = 0 \implies x = -1 \]

### Step 7: Find the Value of \( c \)
Now we have \( x = -1 \) and \( y = 0 \). Substitute these values back into the asymptotic equation:
\[ (-1)^2 + 3(-1)(0) + 2(0)^2 + 2(-1) + 3(0) + c = 0 \]
This simplifies to:
\[ 1 + 0 + 0 - 2 + 0 + c = 0 \implies -1 + c = 0 \implies c = 1 \]

### Step 8: Write the Asymptotic Equation
Now we can write the asymptotic equation:
\[ x^2 + 3xy + 2y^2 + 2x + 3y + 1 = 0 \]

### Step 9: Find the Conjugate Hyperbola Equation
Using the property:
\[ 2 \times (x^2 + 3xy + 2y^2 + 2x + 3y + 1) = (x^2 + 3xy + 2y^2 + 2x + 3y) + \text{(conjugate hyperbola)} \]
This gives us:
\[ 2x^2 + 6xy + 4y^2 + 4x + 6y + 2 = x^2 + 3xy + 2y^2 + 2x + 3y + \text{(conjugate hyperbola)} \]
Thus, the conjugate hyperbola equation is:
\[ x^2 + 3xy + 2y^2 + 2x + 3y + 2 = 0 \]

### Final Answer
The equation of the hyperbola conjugate to the given hyperbola is:
\[ x^2 + 3xy + 2y^2 + 2x + 3y + 2 = 0 \]