Question

If the slope of one line of the pair of lines \( 2x^2 + hxy + 6y^2 = 0 \) is thrice the slope of the other line, then \( h \) = ? 

• A. \( \pm 16 \)
• B. \( \pm 9 \)
• C. \( \pm 18 \)
• D. \( \pm 8 \)

Hint:

When dealing with a pair of lines given by a quadratic equation, use Vieta’s formulas to relate the sum and product of the slopes to the coefficients of the equation.

Answer 1

The Correct Option is D

Step 1: Standard Form of Pair of Lines
The general equation of a pair of straight lines is given by: \[ ax^2 + 2hxy + by^2 = 0. \] Comparing with the given equation: \[ 2x^2 + hxy + 6y^2 = 0, \] we identify: \[ a = 2, \quad 2h = h, \quad b = 6. \] Step 2: Condition for the Slopes
The slopes of the lines are the roots of the equation: \[ m^2 - \frac{-h}{6} m + \frac{2}{6} = 0. \] Simplifying: \[ m^2 + \frac{h}{6} m + \frac{1}{3} = 0. \] Given that one root is three times the other, let the roots be \( m \) and \( 3m \). Using Vieta's formulas: 1. Sum of roots: \[ m + 3m = -\frac{h}{6} \Rightarrow 4m = -\frac{h}{6}. \] \[ m = -\frac{h}{24}. \] 2. Product of roots: \[ m(3m) = \frac{1}{3} \Rightarrow 3m^2 = \frac{1}{3}. \] \[ m^2 = \frac{1}{9}. \] Step 3: Solve for \( h \)
From \( m = -\frac{h}{24} \), squaring both sides: \[ \left( \frac{h}{24} \right)^2 = \frac{1}{9}. \] \[ \frac{h^2}{576} = \frac{1}{9}. \] Multiplying by 576: \[ h^2 = \frac{576}{9} = 64. \] \[ h = \pm 8. \] Final Answer: \[ \boxed{\pm 8}. \]

Answer 2

Given the equation of pair of lines \(2x^2 + hxy + 6y^2 = 0\), let the slopes of the lines be \(m_1\) and \(m_2\). The condition states that \(m_1 = 3m_2\).

For a general second-degree equation of form \(ax^2 + 2hxy + by^2 = 0\), the lines are given by the equation \((x - m_1 y)(x - m_2 y) = 0\). Here, comparing with \(2x^2 + hxy + 6y^2 = 0\), we have \(a = 2\), \(2h = h\), and \(b = 6\).

The slopes are the roots of the quadratic equation:

\(ay^2 + 2hxy + bx^2 = 0 \rightarrow 2 + h \cdot m + 6m^2 = 0\).

Using \(m_1 = 3m_2\), substitute into the slope formula:

\(m_1 \cdot m_2 = \frac{b}{a} = \frac{6}{2} = 3.\)

Thus, \((3m_2) \cdot m_2 = 3 \) ⟹ \(3m_2^2 = 3\) ⟹ \(m_2^2 = 1 \) ⟹ \(m_2 = \pm 1.\)

For \(m_2 = 1\), \(m_1 = 3\). Substituting into slope equation:

\(6(1)^2 + h(1) + 2 = 0\) ⟹ \(6 + h + 2 = 0\) ⟹ \(h = -8\).

For \(m_2 = -1\), \(m_1 = -3\). Substitute and solve:

\(6(-1)^2 + h(-1) + 2 = 0\) ⟹ \(6 - h + 2 = 0\) ⟹ \(-h = -8\) ⟹ \(h = 8\).

Thus, \(h = \pm 8\).