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: Recall slope relations for pair of lines.
For $ax^2+2hxy+by^2=0$: $m_1+m_2=-2h/b$ and $m_1m_2=a/b$.
Step 2: Apply the product condition.
$m_1m_2=2 \implies a/b=2 \implies a=2b$.
Step 3: Apply the difference condition.
$3(m_1-m_2)-7=0 \implies m_1-m_2=7/3$. Using $(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$: \[\frac{49}{9}=\frac{4h^2}{b^2}-8 \implies \frac{4h^2}{b^2}=\frac{121}{9} \implies \frac{h}{b}=\pm\frac{11}{6}.\]
Step 4: Form the ratio $a:b:h$.
With $a=2b$ and $h=\pm 11b/6$, multiplying by 6 gives $6a=12b$ and $6h=\pm 11b$, so \[a:b:h=12:6:(\pm 11) \implies \frac{a}{12}=\frac{b}{6}=\frac{h}{\pm 11}.\]
Step 5: Verify with $b=6,a=12,h=11$.
$m_1m_2=12/6=2$ and $(m_1-m_2)^2=(22/6)^2-8=196/36$ giving $m_1-m_2=7/3$ and $3(m_1-m_2)=7$. Both satisfied.
Step 6: State the answer.
\[ \boxed{\frac{a}{12}=\frac{b}{6}=\frac{h}{\pm 11}} \]