If one of the lines of `ax^(2)+2hxy+by^(2)=0` bisects the angle between the axes, in the first quadrant, then

Options

A. `h^(2)-ab=0`
B. `h^(2)+ab=0`
C. `(a+b)^(2)=h^(2)`
D. `(a+b)^(2)=4h^(2)`

Answer

To solve the problem, we will analyze the given equation of the pair of straight lines and apply the conditions provided in the question.

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
The equation given is \( ax^2 + 2hxy + by^2 = 0 \). This represents a pair of straight lines passing through the origin.

2. **Identifying the Slopes**:
For the pair of lines represented by the equation, we know the slopes \( m_1 \) and \( m_2 \) can be derived from the relationships:
\[ m_1 + m_2 = -\frac{2h}{b} \quad \text{(1)} \]
\[ m_1 \cdot m_2 = \frac{a}{b} \quad \text{(2)} \]

3. **Condition of Angle Bisector**:
The problem states that one of the lines bisects the angle between the axes in the first quadrant. The angle bisector of the axes \( y = x \) has a slope \( m_1 = 1 \).

4. **Substituting the Slope**:
Substitute \( m_1 = 1 \) into equation (1):
\[ 1 + m_2 = -\frac{2h}{b} \]
Rearranging gives:
\[ m_2 = -\frac{2h}{b} - 1 \quad \text{(3)} \]

5. **Using the Product of Slopes**:
Now substitute \( m_1 = 1 \) into equation (2):
\[ 1 \cdot m_2 = \frac{a}{b} \]
Thus, we have:
\[ m_2 = \frac{a}{b} \quad \text{(4)} \]

6. **Equating the Two Expressions for \( m_2 \)**:
From equations (3) and (4), we can equate the two expressions for \( m_2 \):
\[ -\frac{2h}{b} - 1 = \frac{a}{b} \]
Multiplying through by \( b \) to eliminate the denominator:
\[ -2h - b = a \]
Rearranging gives:
\[ a + b + 2h = 0 \quad \text{(5)} \]

7. **Squaring the Result**:
Now, we can square both sides of equation (5):
\[ (a + b)^2 = (2h)^2 \]
This simplifies to:
\[ (a + b)^2 = 4h^2 \]

### Conclusion:
The final result shows that if one of the lines of the equation bisects the angle between the axes in the first quadrant, then:
\[ (a + b)^2 = 4h^2 \]
Thus, the correct option is the fourth one.