For hyperbola whose center is at (1, 2) and the asymptotes are parallel to lines `2x+3y=0` and `x+2y=1` , the equation of the hyperbola passing through (2, 4) is (a) `(2x+3y-5)(x+2y-8)=40` (b) `(2x+3y-8)(x+2y-8)=40` (c) `(2x+3y-8)(x+2y-5)=30` (d) none of these

Options

A. `(2x+3y-5)(x+2y-8)=40`
B. `(2x+3y-8)(x+2y-5)=40`
C. `(2x+3y-8)(x+2y-5)=30`
D. none of these

Explanation

To find the equation of the hyperbola with the given conditions, we will follow these steps:

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

### Step 2: Find the equations of the asymptotes passing through the center
The center of the hyperbola is at \((1, 2)\). We can express the asymptotes in the form:
1. \(2x + 3y + \lambda_1 = 0\)
2. \(x + 2y + \lambda_2 = 0\)

Since the center \((1, 2)\) lies on both asymptotes, we can substitute these coordinates into the equations to find \(\lambda_1\) and \(\lambda_2\).

### Step 3: Substitute the center into the asymptote equations
For the first asymptote:
\[ 2(1) + 3(2) + \lambda_1 = 0 \implies 2 + 6 + \lambda_1 = 0 \implies \lambda_1 = -8 \]
Thus, the first asymptote becomes:
\[ 2x + 3y - 8 = 0 \]

For the second asymptote:
\[ 1 + 2(2) + \lambda_2 = 0 \implies 1 + 4 + \lambda_2 = 0 \implies \lambda_2 = -5 \]
Thus, the second asymptote becomes:
\[ x + 2y - 5 = 0 \]

### Step 4: Write the combined equation for the hyperbola
The equation of the hyperbola can be expressed as the product of the equations of the asymptotes plus a constant \(k\):
\[ (2x + 3y - 8)(x + 2y - 5) = k \]

### Step 5: Use the point (2, 4) to find \(k\)
Substituting the point \((2, 4)\) into the equation:
\[ (2(2) + 3(4) - 8)(2 + 2(4) - 5) = k \]
Calculating each term:
\[ (4 + 12 - 8)(2 + 8 - 5) = k \implies (8)(5) = k \implies k = 40 \]

### Step 6: Write the final equation of the hyperbola
Thus, the equation of the hyperbola is:
\[ (2x + 3y - 8)(x + 2y - 5) = 40 \]

### Conclusion
The correct answer from the options provided is:
**(c) \((2x + 3y - 8)(x + 2y - 5) = 40\)**