Question

If slope of two lines represented by \( \dfrac{x^2}{a}+\dfrac{2xy}{h}+\dfrac{y^2}{b}=0 \) is twice the other, then \(ab:h^2=\ ?\)

• A. \(1:2\)
• B. \(2:1\)
• C. \(8:9\)
• D. \(9:8\)

Hint:

For pair of lines: \[ Ax^2+2Hxy+By^2=0 \] remember: \[ m_1+m_2=-\frac{2H}{B},\qquad m_1m_2=\frac{A}{B} \] These formulas solve most slope-ratio problems quickly.

Answer 1

The Correct Option is D

Concept: The homogeneous second-degree equation \[ Ax^2+2Hxy+By^2=0 \] represents a pair of straight lines passing through the origin. If slopes are \(m_1\) and \(m_2\), then: \[ m_1+m_2=-\frac{2H}{B} \] and \[ m_1m_2=\frac{A}{B} \]

Step 1: Comparing with standard equation.
Given: \[ \frac{x^2}{a}+\frac{2xy}{h}+\frac{y^2}{b}=0 \] Comparing with \[ Ax^2+2Hxy+By^2=0 \] we get: \[ A=\frac1a,\qquad H=\frac1h,\qquad B=\frac1b \] Therefore: \[ m_1+m_2=-\frac{2/h}{1/b} =-\frac{2b}{h} \] and \[ m_1m_2=\frac{1/a}{1/b} =\frac{b}{a} \]

Step 2: Using given condition on slopes.
One slope is twice the other. Let: \[ m_1=2m,\qquad m_2=m \] Then: \[ m_1+m_2=3m \] Using sum relation: \[ 3m=-\frac{2b}{h} \] Hence: \[ m=-\frac{2b}{3h} \]

Step 3: Using product relation.
Now: \[ m_1m_2=2m^2 \] But product also equals: \[ \frac{b}{a} \] Therefore: \[ 2m^2=\frac{b}{a} \] Substituting \(m=-\frac{2b}{3h}\): \[ 2\left(\frac{2b}{3h}\right)^2=\frac{b}{a} \] \[ 2\cdot\frac{4b^2}{9h^2}=\frac{b}{a} \] \[ \frac{8b^2}{9h^2}=\frac{b}{a} \] Cross multiplying: \[ 8ab^2=9bh^2 \] Dividing by \(b\): \[ 8ab=9h^2 \] Thus: \[ \frac{ab}{h^2}=\frac98 \] Hence, \[ ab:h^2=9:8 \]