The equation of the hyperbola whose asymptotes are 3x+4y-2=0, 2x+y+1=0 and which passes through the point (1, 1) is

Options

A. `6x^(2)+41xy-44y^(2)-30x+2y-22=0`
B. `6x^(2)+11xy+4y^(2)-x+2y-22=0`
C. `6x^(2)-15xy+14y^(2)-6x+12y-12=0`
D. `6x^(2)+13xy+6y^(2)-38x-37y-98=0`

Answer

To find the equation of the hyperbola whose asymptotes are given and which passes through the point (1, 1), we can follow these steps:

### Step 1: Identify the equations of the asymptotes
The equations of the asymptotes are given as:
1. \(3x + 4y - 2 = 0\)
2. \(2x + y + 1 = 0\)

### Step 2: Form the equation of the pair of asymptotes
To find the equation of the pair of asymptotes, we multiply the two equations:
\[ (3x + 4y - 2)(2x + y + 1) = 0 \]

### Step 3: Expand the product
Expanding the product, we have:
\[ 3x(2x) + 3x(y) + 3x(1) + 4y(2x) + 4y(y) + 4y(1) - 2(2x) - 2(y) - 2(1) = 0 \]
This simplifies to:
\[ 6x^2 + 3xy + 3x + 8xy + 4y^2 + 4y - 4x - 2y - 2 = 0 \]

### Step 4: Combine like terms
Combining like terms gives us:
\[ 6x^2 + (3xy + 8xy) + (3x - 4x) + 4y^2 + (4y - 2y) - 2 = 0 \]
This simplifies to:
\[ 6x^2 + 11xy + 4y^2 - x + 2y - 2 = 0 \]

### Step 5: Write the general form of the hyperbola
The equation of the hyperbola can be expressed as:
\[ 6x^2 + 4y^2 + 11xy - x + 2y - \lambda = 0 \]
where \(\lambda\) is a constant that we need to determine.

### Step 6: Substitute the point (1, 1) into the equation
Since the hyperbola passes through the point (1, 1), we substitute \(x = 1\) and \(y = 1\):
\[ 6(1)^2 + 4(1)^2 + 11(1)(1) - (1) + 2(1) - \lambda = 0 \]
This simplifies to:
\[ 6 + 4 + 11 - 1 + 2 - \lambda = 0 \]
\[ 22 - \lambda = 0 \]
Thus, \(\lambda = 22\).

### Step 7: Substitute \(\lambda\) back into the hyperbola equation
Now, we substitute \(\lambda = 22\) back into the hyperbola equation:
\[ 6x^2 + 4y^2 + 11xy - x + 2y - 22 = 0 \]

### Final Answer
The equation of the hyperbola is:
\[ 6x^2 + 4y^2 + 11xy - x + 2y - 22 = 0 \]

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