The equation of hyperbola whose asymtotes are the straight lines `3x-4y+7=0" and "4x+3y+1=0` and which passes through the origin is

Options

A. `(3x-4y+7)(4x+3y+1)+7=0`
B. `(3x-4y+7)(4x+3y+1)-7=0`
C. `(3x-4y+7)(4x+3y-1)+7=0`
D. `(3x-4y+7)(4x-3y+1)=0`

Answer

To find the equation of the hyperbola whose asymptotes are given by the lines \(3x - 4y + 7 = 0\) and \(4x + 3y + 1 = 0\), and which passes through the origin, we can follow these steps:

### Step 1: Identify the Asymptotes
The asymptotes of the hyperbola are given as:
1. \(l_1: 3x - 4y + 7 = 0\)
2. \(l_2: 4x + 3y + 1 = 0\)

### Step 2: Write the Equation of the Hyperbola
The general form of the hyperbola with asymptotes \(l_1\) and \(l_2\) can be expressed as:
\[ l_1 \cdot l_2 + k = 0 \]
Substituting the equations of the asymptotes into this form gives:
\[ (3x - 4y + 7)(4x + 3y + 1) + k = 0 \]

### Step 3: Expand the Product
Now, we need to expand the product:
\[ (3x - 4y + 7)(4x + 3y + 1) \]
Using the distributive property (FOIL), we get:
\[ = 3x(4x) + 3x(3y) + 3x(1) - 4y(4x) - 4y(3y) - 4y(1) + 7(4x) + 7(3y) + 7(1) \]
Calculating each term:
\[ = 12x^2 + 9xy + 3x - 16xy - 12y^2 - 4y + 28x + 21y + 7 \]
Combining like terms:
\[ = 12x^2 + (9xy - 16xy) + (3x + 28x) + (-12y^2) + (-4y + 21y) + 7 \]
This simplifies to:
\[ = 12x^2 - 7xy - 12y^2 + 31x + 17y + 7 \]

### Step 4: Substitute the Point (0, 0)
Since the hyperbola passes through the origin (0, 0), we substitute \(x = 0\) and \(y = 0\) into the equation:
\[ 12(0)^2 - 7(0)(0) - 12(0)^2 + 31(0) + 17(0) + 7 + k = 0 \]
This simplifies to:
\[ 7 + k = 0 \]
From which we find:
\[ k = -7 \]

### Step 5: Write the Final Equation
Substituting \(k = -7\) back into the equation gives:
\[ 12x^2 - 7xy - 12y^2 + 31x + 17y + 7 - 7 = 0 \]
Thus, the equation simplifies to:
\[ 12x^2 - 7xy - 12y^2 + 31x + 17y = 0 \]

### Final Answer
The equation of the hyperbola is:
\[ 12x^2 - 7xy - 12y^2 + 31x + 17y = 0 \]