Question

Let the slopes $ m_1 $ and $ m_2 $ of the lines represented by: $$ ax^2 + 2hxy + by^2 = 0 $$ satisfy: $$ 3(m_1 - m_2) - 7 = 0 \quad \text{and} \quad m_1 m_2 - 2 = 0 $$ Then find the correct relation among $ a, b, h $.

• A. \( \frac{a}{12} = \frac{b}{6} = \frac{h}{\pm 11} \)
• B. \( \frac{a}{6} = \frac{b}{12} = \frac{h}{\pm 11} \)
• C. \( a = b = \pm h \)
• D. \( \frac{a}{2} = b = \pm h \)

Hint:

Use root properties of the quadratic formed by slope equation from homogeneous second degree to relate coefficients.

Answer 1

The Correct Option is A

The homogeneous equation of second degree: \[ ax^2 + 2hxy + by^2 = 0 \] represents two straight lines passing through the origin. The slopes of the lines are roots of the equation: \[ am^2 + 2hm + b = 0 \] Let roots be \( m_1, m_2 \). Then,
- Sum of roots: \( m_1 + m_2 = \frac{-2h}{a} \)
- Product of roots: \( m_1 m_2 = \frac{b}{a} \) Given: \[ 3(m_1 - m_2) = 7 \Rightarrow m_1 - m_2 = \frac{7}{3} \quad \text{and} \quad m_1 m_2 = 2 \] We also know: \[ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 \Rightarrow \left(\frac{7}{3}\right)^2 = (m_1 + m_2)^2 - 8 \Rightarrow \frac{49}{9} = (m_1 + m_2)^2 - 8 \Rightarrow (m_1 + m_2)^2 = \frac{121}{9} \Rightarrow m_1 + m_2 = \pm \frac{11}{3} \] Now: \[ m_1 + m_2 = \frac{-2h}{a} = \pm \frac{11}{3} \Rightarrow h = \mp \frac{11a}{6} \] Also: \[ m_1 m_2 = \frac{b}{a} = 2 \Rightarrow b = 2a \] Now, plug into:
- \( a = a \)
- \( b = 2a \)
- \( h = \mp \frac{11a}{6} \) Divide all by \( a \): \[ \frac{a}{12} = \frac{b}{6} = \frac{h}{\pm 11} \Rightarrow \boxed{ \frac{a}{12} = \frac{b}{6} = \frac{h}{\pm 11} } \]