Step 1: Understanding the Concept:
The given equation is a homogeneous second-degree equation in \( x \) and \( y \).
Such equations always represent a pair of straight lines passing through the origin.
The slopes of these lines, \( m_1 \) and \( m_2 \), are related to the coefficients of the equation.
We use the sum of slopes and product of slopes formulas derived from the general form \( Ax^2 + 2Hxy + By^2 = 0 \).
Step 2: Key Formula or Approach:
For \( Ax^2 + 2Hxy + By^2 = 0 \):
Sum of slopes \( m_1 + m_2 = - \frac{2H}{B} \)
Product of slopes \( m_1 m_2 = \frac{A}{B} \)
In our problem, comparing with the standard form:
\( A = 1/a \), \( 2H = 2/h \implies H = 1/h \), \( B = 1/b \).
Step 3: Detailed Explanation:
Let the two slopes be \( m_1 \) and \( m_2 \). We are given \( m_1 = 2m_2 \).
Sum of slopes:
\[ m_1 + m_2 = - \frac{2/h}{1/b} = - \frac{2b}{h} \]
Substituting \( m_1 = 2m_2 \):
\[ 2m_2 + m_2 = - \frac{2b}{h} \implies 3m_2 = - \frac{2b}{h} \implies m_2 = - \frac{2b}{3h} \]
Product of slopes:
\[ m_1 \cdot m_2 = \frac{1/a}{1/b} = \frac{b}{a} \]
Substituting \( m_1 = 2m_2 \):
\[ 2m_2^2 = \frac{b}{a} \]
Now substitute the expression for \( m_2 \) from the sum formula:
\[ 2 \left( - \frac{2b}{3h} \right)^2 = \frac{b}{a} \]
\[ 2 \left( \frac{4b^2}{9h^2} \right) = \frac{b}{a} \]
\[ \frac{8b^2}{9h^2} = \frac{b}{a} \]
Dividing by \( b \) on both sides (assuming \( b \neq 0 \)):
\[ \frac{8b}{9h^2} = \frac{1}{a} \implies 8ab = 9h^2 \]
To find the ratio \( ab : h^2 \):
\[ \frac{ab}{h^2} = \frac{9}{8} \]
Step 4: Final Answer:
The ratio \( ab : h^2 \) is \( 9 : 8 \).
This corresponds to option (D).