Question

Suppose the slopes \(m_1\) and \(m_2\) of the lines represented by \[ ax^2+2hxy+by^2=0 \] satisfy \[ 3(m_1-m_2)-7=0 \] and \[ m_1m_2-2=0. \] Then which of the following is true?

• A. \(\dfrac{a}{12}=\dfrac{b}{6}=\dfrac{h}{\pm11}\)
• B. \(\dfrac{a}{6}=\dfrac{b}{12}=\dfrac{h}{\pm11}\)
• C. \(a=b=\pm h\)
• D. \(\dfrac{a}{2}=b=\pm h\)

Hint:

For the pair of lines \[ ax^2+2hxy+by^2=0, \] the slopes satisfy: \[ m_1+m_2=-\frac{2h}{b}, \qquad m_1m_2=\frac{a}{b}. \] Use these relations directly in slope-based questions.

Answer 1

The Correct Option is A

Step 1: Use the equation for slopes of pair of lines.
For the homogeneous equation \[ ax^2+2hxy+by^2=0, \] the slopes \(m_1,m_2\) satisfy \[ bm^2+2hm+a=0 \] Therefore, \[ m_1+m_2=-\frac{2h}{b} \] and \[ m_1m_2=\frac{a}{b} \]

Step 2: Use the given product condition.
Given, \[ m_1m_2-2=0 \] So, \[ m_1m_2=2 \] Hence, \[ \frac{a}{b}=2 \] Therefore, \[ a=2b \]

Step 3: Use the given difference condition.
Given, \[ 3(m_1-m_2)-7=0 \] \[ m_1-m_2=\frac73 \] Now, \[ (m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2 \] Substitute values: \[ \left(\frac73\right)^2=(m_1+m_2)^2-4(2) \] \[ \frac{49}{9}=(m_1+m_2)^2-8 \] \[ (m_1+m_2)^2=\frac{49}{9}+\frac{72}{9} \] \[ (m_1+m_2)^2=\frac{121}{9} \] \[ m_1+m_2=\pm\frac{11}{3} \]

Step 4: Relate this to coefficients.
Since \[ m_1+m_2=-\frac{2h}{b}, \] we get \[ -\frac{2h}{b}=\pm\frac{11}{3} \] Thus, \[ h=\mp\frac{11b}{6} \] Also, \[ a=2b \]

Step 5: Express in proportional form.
Take \[ b=6k \] Then, \[ a=12k \] and \[ h=\pm11k \] Hence, \[ \frac{a}{12}=\frac{b}{6}=\frac{h}{\pm11} \]

Step 6: Final conclusion.
Therefore, \[ \boxed{ \frac{a}{12}=\frac{b}{6}=\frac{h}{\pm11} } \]