The equation `ax^(2)+2hxy+ay^(2)=0` represents a pair of coincident lines through origin, if

Options

A. `h=2a`
B. `2h=a`
C. `h=pma`
D. `2h^(2)=a`

Answer

To determine the condition under which the equation \( ax^2 + 2hxy + by^2 = 0 \) represents a pair of coincident lines through the origin, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Form**: The given equation is \( ax^2 + 2hxy + by^2 = 0 \). This is a homogeneous equation representing a pair of straight lines through the origin.

2. **Condition for Coincident Lines**: For the lines represented by the equation to be coincident, the angle \( \theta \) between the lines must be either \( 0^\circ \) or \( 180^\circ \). This means that the tangent of the angle \( \theta \) must be equal to 0.

3. **Use the Formula for Tangent of Angle**: The formula for the tangent of the angle between the lines represented by the equation is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
For the lines to be coincident, we set \( \tan \theta = 0 \).

4. **Set the Tangent Equal to Zero**: Setting the formula for \( \tan \theta \) to zero gives us:
\[ \frac{2\sqrt{h^2 - ab}}{a + b} = 0 \]
This implies that the numerator must be zero (since the denominator cannot be zero for the equation to hold).

5. **Solve the Numerator**: Thus, we have:
\[ 2\sqrt{h^2 - ab} = 0 \]
Dividing both sides by 2 gives:
\[ \sqrt{h^2 - ab} = 0 \]

6. **Square Both Sides**: Squaring both sides results in:
\[ h^2 - ab = 0 \]
Therefore, we can conclude that:
\[ h^2 = ab \]

### Conclusion:
The equation \( ax^2 + 2hxy + by^2 = 0 \) represents a pair of coincident lines through the origin if and only if:
\[ h^2 = ab \]