Question

Find \( k \), if the sum of the slopes of the lines represented by \[ x^2 + kxy - 3y^2 = 0 \] is twice their product.

Hint:

For homogeneous second-degree equations, the slopes satisfy: \[ m_1 + m_2 = -\frac{2H}{B}, \quad m_1 m_2 = \frac{A}{B}. \]

Answer 1

Step 1: Determine the Slopes of the Lines
The general form of a homogeneous second-degree equation: \[ Ax^2 + 2Hxy + By^2 = 0 \] represents two straight lines. The slopes of these lines are given by: \[ m_1, m_2 = \frac{- (H \pm \sqrt{H^2 - AB})}{B}. \] By comparing this with the equation \( x^2 + kxy - 3y^2 = 0 \), we identify the coefficients: \[ A = 1, \quad H = \frac{k}{2}, \quad B = -3. \] Step 2: Apply the Given Condition
The sum of the slopes is: \[ m_1 + m_2 = -\frac{2H}{B} = -\frac{k}{-3} = \frac{k}{3}. \] The product of the slopes is: \[ m_1 m_2 = \frac{A}{B} = \frac{1}{-3} = -\frac{1}{3}. \] Using the condition \( m_1 + m_2 = 2(m_1 m_2) \), we substitute the known values: \[ \frac{k}{3} = 2 \times \left(-\frac{1}{3}\right). \] \[ \frac{k}{3} = -\frac{2}{3}. \] Step 3: Solve for \( k \)
\[ k = -2. \]