Question

For \(a,b,h>0\), if the slope of one of the lines represented by \[ a^2x^2+2hxy+b^2y^2=0 \] is twice that of the other, then the value of \[ \frac{h}{ab} \] is:

• A. \[ \frac{3\sqrt{2}}{4} \]
• B. \[ \frac{2\sqrt{3}}{4} \]
• C. \[ -\frac{2\sqrt{3}}{4} \]
• D. \[ -\frac{3\sqrt{2}}{4} \]

Hint:

For a homogeneous equation \(Ax^2+2Hxy+By^2=0\), the slopes of the two lines are obtained from \(Bm^2+2Hm+A=0\). Then use sum and product of roots.

Answer 1

The Correct Option is A

Step 1: Find the slopes of the pair of lines.
The homogeneous second-degree equation \[ a^2x^2+2hxy+b^2y^2=0 \] represents two straight lines through the origin.
Let their slopes be \(m_1\) and \(m_2\).
Putting \[ y=mx, \] we get \[ a^2+2hm+b^2m^2=0. \] Thus the slopes satisfy \[ b^2m^2+2hm+a^2=0. \]

Step 2: Use the condition on the slopes.
Given that one slope is twice the other.
Let \[ m_1=2m,\qquad m_2=m. \] Using the product of roots, \[ m_1m_2=\frac{a^2}{b^2}. \] Hence, \[ 2m^2=\frac{a^2}{b^2}. \] \[ m=\frac{a}{b\sqrt2}. \]

Step 3: Use the sum of roots.
The sum of roots is \[ m_1+m_2 = -\frac{2h}{b^2}. \] Substituting \(m_1=2m\) and \(m_2=m\), \[ 3m = -\frac{2h}{b^2}. \] Taking magnitude (since \(h>0\)), \[ 3\left(\frac{a}{b\sqrt2}\right) = \frac{2h}{b^2}. \] Multiplying by \(b^2\), \[ \frac{3ab}{\sqrt2} = 2h. \] Therefore, \[ h=\frac{3ab}{2\sqrt2}. \]

Step 4: Compute \(\frac{h}{ab}\).
\[ \frac{h}{ab} = \frac{3}{2\sqrt2}. \] Rationalizing, \[ \frac{h}{ab} = \frac{3\sqrt2}{4}. \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{h}{ab}=\frac{3\sqrt2}{4}} \]