The asymptotes of the hyperbola ` 6x^(2) +13xy +6y^(2) -7x -8y-26 =0` are

Options

A. A:2x+3y-1=0, 3x+2y+2=0
B. B:2x+3y=1, 3x+2y=2
C. C:2x+3y=0, 3x+2=0
D. D:2x+3y=3, 3x+2y=4

Answer

To find the asymptotes of the hyperbola given by the equation \( 6x^2 + 13xy + 6y^2 - 7x - 8y - 26 = 0 \), we can follow these steps:

### Step 1: Write the equation of the hyperbola
The equation of the hyperbola is given as:
\[ 6x^2 + 13xy + 6y^2 - 7x - 8y - 26 = 0 \]

### Step 2: Formulate the equation for the asymptotes
The equations of the asymptotes can be found by adding a constant \( \lambda \) to the hyperbola equation:
\[ 6x^2 + 13xy + 6y^2 - 7x - 8y - 26 + \lambda = 0 \]

### Step 3: Find the center of the hyperbola
To find the center, we need to partially differentiate the equation with respect to \( x \) and \( y \) and set the derivatives to zero.

1. **Partial differentiation with respect to \( x \)**:
\[ \frac{\partial}{\partial x}(6x^2 + 13xy + 6y^2 - 7x - 8y - 26) = 12x + 13y - 7 = 0 \quad \text{(Equation 1)} \]

2. **Partial differentiation with respect to \( y \)**:
\[ \frac{\partial}{\partial y}(6x^2 + 13xy + 6y^2 - 7x - 8y - 26) = 13x + 12y - 8 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we solve the two equations obtained from partial differentiation:

From Equation 1:
\[ 12x + 13y = 7 \quad \text{(1)} \]

From Equation 2:
\[ 13x + 12y = 8 \quad \text{(2)} \]

We can solve these equations simultaneously.

Multiply Equation (1) by 13:
\[ 156x + 169y = 91 \quad \text{(3)} \]

Multiply Equation (2) by 12:
\[ 156x + 144y = 96 \quad \text{(4)} \]

Now subtract Equation (4) from Equation (3):
\[ (169y - 144y) = 91 - 96 \\ 25y = -5 \\ y = -\frac{1}{5} \]

Substituting \( y = -\frac{1}{5} \) back into Equation (1):
\[ 12x + 13\left(-\frac{1}{5}\right) = 7 \\ 12x - \frac{13}{5} = 7 \\ 12x = 7 + \frac{13}{5} \\ 12x = \frac{35}{5} + \frac{13}{5} = \frac{48}{5} \\ x = \frac{48}{60} = \frac{4}{5} \]

Thus, the center of the hyperbola is:
\[ \left(\frac{4}{5}, -\frac{1}{5}\right) \]

### Step 5: Substitute the center into the asymptote equation
Now, substitute \( x = \frac{4}{5} \) and \( y = -\frac{1}{5} \) into the modified asymptote equation:
\[ 6\left(\frac{4}{5}\right)^2 + 13\left(\frac{4}{5}\right)\left(-\frac{1}{5}\right) + 6\left(-\frac{1}{5}\right)^2 - 7\left(\frac{4}{5}\right) - 8\left(-\frac{1}{5}\right) - 26 + \lambda = 0 \]

Calculating each term:
\[ = 6\left(\frac{16}{25}\right) - \frac{52}{25} + 6\left(\frac{1}{25}\right) - \frac{28}{5} + \frac{8}{5} - 26 + \lambda = 0 \]
\[ = \frac{96}{25} - \frac{52}{25} + \frac{6}{25} - \frac{140}{25} + \frac{40}{25} - \frac{650}{25} + \lambda = 0 \]
Combining terms:
\[ \frac{96 - 52 + 6 - 140 + 40 - 650}{25} + \lambda = 0 \]
\[ \frac{-700}{25} + \lambda = 0 \\ \lambda = 28 \]

### Step 6: Write the final equation of the asymptotes
Substituting \( \lambda \) back into the asymptote equation:
\[ 6x^2 + 13xy + 6y^2 - 7x - 8y - 26 + 28 = 0 \]
This simplifies to:
\[ 6x^2 + 13xy + 6y^2 - 7x - 8y + 2 = 0 \]

### Step 7: Conclusion
The asymptotes of the hyperbola are given by the equation:
\[ 6x^2 + 13xy + 6y^2 - 7x - 8y + 2 = 0 \]