If the slope of one of the lines given by `6x^(2)+axy+y^(2)=0` exceeds the slope of the other by one, then a is equal to

Options

A. `pm 2`
B. 5
C. `-5`
D. `pm 5`

Answer

To solve the problem, we need to find the value of \( a \) such that the slopes of the lines given by the equation \( 6x^2 + axy + y^2 = 0 \) differ by 1. Here’s a step-by-step solution:

### Step 1: Identify the form of the equation
The given equation is a quadratic in \( x \) and \( y \):
\[ 6x^2 + axy + y^2 = 0 \]
This can be rewritten in the form \( Ax^2 + Bxy + Cy^2 = 0 \) where \( A = 6 \), \( B = a \), and \( C = 1 \).

### Step 2: Use the formula for the slopes of the lines
The slopes \( m_1 \) and \( m_2 \) of the lines represented by the equation can be found using the formula:
\[ m_1 + m_2 = -\frac{B}{A} = -\frac{a}{6} \]
\[ m_1 m_2 = \frac{C}{A} = \frac{1}{6} \]

### Step 3: Set up the equations based on the given condition
We know from the problem statement that the slope of one line exceeds the slope of the other by 1:
\[ m_1 = m_2 + 1 \]

### Step 4: Substitute \( m_1 \) in terms of \( m_2 \)
Substituting \( m_1 \) in the sum of slopes equation:
\[ (m_2 + 1) + m_2 = -\frac{a}{6} \]
This simplifies to:
\[ 2m_2 + 1 = -\frac{a}{6} \]
Thus, we can express \( m_2 \):
\[ 2m_2 = -\frac{a}{6} - 1 \quad \Rightarrow \quad m_2 = -\frac{a + 6}{12} \]

### Step 5: Substitute \( m_2 \) into the product of slopes equation
Now substitute \( m_2 \) into the product of slopes equation:
\[ m_1 m_2 = \frac{1}{6} \]
Substituting \( m_1 = m_2 + 1 \):
\[ (m_2 + 1) m_2 = \frac{1}{6} \]
Substituting \( m_2 = -\frac{a + 6}{12} \):
\[ \left(-\frac{a + 6}{12} + 1\right)\left(-\frac{a + 6}{12}\right) = \frac{1}{6} \]

### Step 6: Solve the equation
Let’s simplify:
\[ \left(-\frac{a + 6 - 12}{12}\right)\left(-\frac{a + 6}{12}\right) = \frac{1}{6} \]
This simplifies to:
\[ \left(-\frac{a - 6}{12}\right)\left(-\frac{a + 6}{12}\right) = \frac{1}{6} \]
\[ \frac{(a - 6)(a + 6)}{144} = \frac{1}{6} \]
Cross-multiplying gives:
\[ (a^2 - 36) = 24 \quad \Rightarrow \quad a^2 = 60 \quad \Rightarrow \quad a = \pm \sqrt{60} = \pm 2\sqrt{15} \]

### Step 7: Conclusion
Thus, the possible values for \( a \) are \( 2\sqrt{15} \) and \( -2\sqrt{15} \).