If the slopes of the lines represented by `3x^(2)+kxy-y^(2)=0` differ by 4, then k=

Options

A. `2`
B. `-2`
C. `pm2`
D. `4`

Answer

To solve the problem, we need to find the value of \( k \) such that the slopes of the lines represented by the equation \( 3x^2 + kxy - y^2 = 0 \) differ by 4.

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ 3x^2 + kxy - y^2 = 0 \]

2. **Divide the entire equation by \( x^2 \)** to simplify it:
\[ 3 + k \frac{y}{x} - \left(\frac{y}{x}\right)^2 = 0 \]
Let \( m = \frac{y}{x} \) (the slope), then we can rewrite the equation as:
\[ -m^2 + km + 3 = 0 \]

3. **Rearranging gives us a quadratic equation in terms of \( m \)**:
\[ m^2 - km - 3 = 0 \]

4. **Using the quadratic formula** to find the slopes \( m_1 \) and \( m_2 \):
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -k \), and \( c = -3 \):
\[ m = \frac{k \pm \sqrt{k^2 + 12}}{2} \]

5. **Let \( m_1 \) and \( m_2 \) be the two slopes**:
\[ m_1 = \frac{k + \sqrt{k^2 + 12}}{2}, \quad m_2 = \frac{k - \sqrt{k^2 + 12}}{2} \]

6. **The difference between the slopes** is given as:
\[ m_1 - m_2 = 4 \]

7. **Calculating the difference**:
\[ m_1 - m_2 = \left(\frac{k + \sqrt{k^2 + 12}}{2}\right) - \left(\frac{k - \sqrt{k^2 + 12}}{2}\right) = \frac{2\sqrt{k^2 + 12}}{2} = \sqrt{k^2 + 12} \]
Setting this equal to 4:
\[ \sqrt{k^2 + 12} = 4 \]

8. **Squaring both sides** to eliminate the square root:
\[ k^2 + 12 = 16 \]

9. **Solving for \( k^2 \)**:
\[ k^2 = 16 - 12 = 4 \]

10. **Taking the square root** gives us:
\[ k = 2 \quad \text{or} \quad k = -2 \]

### Final Answer:
\[ k = 2 \quad \text{or} \quad k = -2 \]