Question:
If the slope of one of the lines is twice the slope of the other in the pair of straight lines
\[
6x^2+2hxy+y^2=0,
\]
then
\[
|h|=
\]
If the slope of one of the lines is twice the slope of the other in the pair of straight lines
\[
6x^2+2hxy+y^2=0,
\]
then
\[
|h|=
\]
Show Hint
For
\[
ax^2+2hxy+by^2=0,
\]
substitute
\[
y=mx
\]
to get the quadratic equation in \(m\). The roots are the slopes of the two lines, so Vieta's formulas can be applied immediately.
Updated On: Jul 29, 2026
- \(-\dfrac{3\sqrt3}{2}\)
- \(\dfrac{3\sqrt2}{3}\)
- \(\dfrac{3\sqrt3}{2}\)
- \(\dfrac{3\sqrt5}{2}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For the homogeneous second-degree equation
\[
ax^2+2hxy+by^2=0,
\]
the slopes \(m\) of the two lines are obtained by putting
\[
y=mx.
\]
This gives the quadratic equation
\[
bm^2+2hm+a=0.
\]
The roots of this equation are the slopes of the two lines.
Step 1: Find the equation whose roots are the slopes. Given \[ 6x^2+2hxy+y^2=0. \] Putting \[ y=mx, \] we get \[ 6x^2+2hmx^2+m^2x^2=0. \] Dividing by \(x^2\), \[ m^2+2hm+6=0. \] Let the slopes be \[ m_1,\;m_2. \] Then \[ m_1+m_2=-2h, \] \[ m_1m_2=6. \]
Step 2: Use the given condition. One slope is twice the other. Let \[ m_1=2m_2. \] Let \[ m_2=k. \] Then \[ m_1=2k. \] Using \[ m_1m_2=6, \] \[ (2k)(k)=6. \] \[ 2k^2=6. \] \[ k^2=3. \] \[ k=\pm\sqrt3. \] Thus the slopes are \[ \sqrt3,\;2\sqrt3 \] or \[ -\sqrt3,\;-2\sqrt3. \]
Step 3: Find \(h\). Using \[ m_1+m_2=-2h, \] \[ 3\sqrt3=-2h \] or \[ -3\sqrt3=-2h. \] Hence, \[ h=\pm\frac{3\sqrt3}{2}. \] Therefore, \[ |h| = \frac{3\sqrt3}{2}. \]
Step 4: Write the final answer. \[ \boxed{\frac{3\sqrt3}{2}} \]
Step 1: Find the equation whose roots are the slopes. Given \[ 6x^2+2hxy+y^2=0. \] Putting \[ y=mx, \] we get \[ 6x^2+2hmx^2+m^2x^2=0. \] Dividing by \(x^2\), \[ m^2+2hm+6=0. \] Let the slopes be \[ m_1,\;m_2. \] Then \[ m_1+m_2=-2h, \] \[ m_1m_2=6. \]
Step 2: Use the given condition. One slope is twice the other. Let \[ m_1=2m_2. \] Let \[ m_2=k. \] Then \[ m_1=2k. \] Using \[ m_1m_2=6, \] \[ (2k)(k)=6. \] \[ 2k^2=6. \] \[ k^2=3. \] \[ k=\pm\sqrt3. \] Thus the slopes are \[ \sqrt3,\;2\sqrt3 \] or \[ -\sqrt3,\;-2\sqrt3. \]
Step 3: Find \(h\). Using \[ m_1+m_2=-2h, \] \[ 3\sqrt3=-2h \] or \[ -3\sqrt3=-2h. \] Hence, \[ h=\pm\frac{3\sqrt3}{2}. \] Therefore, \[ |h| = \frac{3\sqrt3}{2}. \]
Step 4: Write the final answer. \[ \boxed{\frac{3\sqrt3}{2}} \]
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