The asymptotes of a hyperbola are parallel to lines `2x+3y=0 and 3x+2y=0`. The hyperbola has its centre at (1, 2) and it passes through (5, 3). Find its equation.

Answer

To find the equation of the hyperbola given the asymptotes and other conditions, we can follow these steps:

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

### Step 2: Find the center of the hyperbola
The center of the hyperbola is given as \((1, 2)\). This means that the equations of the asymptotes can be rewritten in terms of the center.

### Step 3: Rewrite the equations of the asymptotes
The equations of the asymptotes can be expressed in the form:
- For \(2x + 3y = 0\):
\[ 2(x - 1) + 3(y - 2) = 0 \implies 2x + 3y - 8 = 0 \]

- For \(3x + 2y = 0\):
\[ 3(x - 1) + 2(y - 2) = 0 \implies 3x + 2y - 7 = 0 \]

### Step 4: Set up the equation of the hyperbola
The equation of the hyperbola can be expressed as:
\[ (2x + 3y - 8)(3x + 2y - 7) = k \]
where \(k\) is a constant.

### Step 5: Substitute the point (5, 3) into the equation
Since the hyperbola passes through the point \((5, 3)\), we substitute \(x = 5\) and \(y = 3\) into the equation:
\[ (2(5) + 3(3) - 8)(3(5) + 2(3) - 7) = k \]

Calculating each part:
- First part:
\[ 2(5) + 3(3) - 8 = 10 + 9 - 8 = 11 \]

- Second part:
\[ 3(5) + 2(3) - 7 = 15 + 6 - 7 = 14 \]

Now, substituting these values:
\[ 11 \times 14 = k \implies k = 154 \]

### Step 6: Write the final equation of the hyperbola
Substituting \(k\) back into the hyperbola equation gives:
\[ (2x + 3y - 8)(3x + 2y - 7) = 154 \]

### Final Answer
The equation of the hyperbola is:
\[ (2x + 3y - 8)(3x + 2y - 7) = 154 \]