if `X^2/a+y^2/b+(2xy)/h=0` represent pair of straight lies and slope one line is twice the other line then `ab:h^2`.

Options

A. ` 9 : 8 `
B. ` 8 : 9 `
C. ` 1 : 2 `
D. ` 2 : 1 `

Answer

To solve the problem, we need to analyze the given equation and derive the required ratio \( \frac{ab}{h^2} \).

### Step 1: Understand the given equation
The equation given is:
\[ \frac{x^2}{a} + \frac{y^2}{b} + \frac{2xy}{h} = 0 \]
This represents a pair of straight lines passing through the origin.

### Step 2: Identify coefficients
We can rewrite the equation in the standard form of a pair of straight lines:
\[ Ax^2 + By^2 + 2Hxy = 0 \]
From our equation, we identify:
- \( A = \frac{1}{a} \)
- \( B = \frac{1}{b} \)
- \( H = \frac{1}{h} \)

### Step 3: Use the properties of slopes
For the pair of straight lines:
- The sum of the slopes \( m_1 + m_2 = -\frac{2H}{B} \)
- The product of the slopes \( m_1 m_2 = \frac{A}{B} \)

Substituting our values:
1. \( m_1 + m_2 = -\frac{2 \cdot \frac{1}{h}}{\frac{1}{b}} = -\frac{2b}{h} \)
2. \( m_1 m_2 = \frac{\frac{1}{a}}{\frac{1}{b}} = \frac{b}{a} \)

### Step 4: Set up the slopes
Let:
- \( m_1 = m \)
- \( m_2 = 2m \)

Now substituting into the equations:
1. \( m + 2m = -\frac{2b}{h} \)
\[ 3m = -\frac{2b}{h} \quad \Rightarrow \quad m = -\frac{2b}{3h} \]

2. \( m \cdot 2m = \frac{b}{a} \)
\[ 2m^2 = \frac{b}{a} \]
Substituting \( m \):
\[ 2\left(-\frac{2b}{3h}\right)^2 = \frac{b}{a} \]
\[ 2 \cdot \frac{4b^2}{9h^2} = \frac{b}{a} \]
\[ \frac{8b^2}{9h^2} = \frac{b}{a} \]

### Step 5: Cross-multiply to find the ratio
Cross-multiplying gives:
\[ 8b^2a = 9bh^2 \]
Dividing both sides by \( b \) (assuming \( b \neq 0 \)):
\[ 8ba = 9h^2 \]
Thus, we can express the ratio:
\[ \frac{ab}{h^2} = \frac{9}{8} \]

### Final Result
The required ratio \( \frac{ab}{h^2} \) is:
\[ \frac{ab}{h^2} = \frac{9}{8} \]