If one of the lines represented by the equation `ax^2+2hxy+by^2=0` is coincident with one of the lines represented by `a'x^2+2h'xy+b'y^2=0` , then

Options

A. `(h'b-hb')(ha'-h'a)`
B. `4(h'b-hb')(ha'-h'a)`
C. `2(h'b-hb')(ha'-h'a)`
D. `4(h'b+hb')(ha'+h'a)`

Explanation

To solve the problem, we need to analyze the conditions under which one line represented by the equation \( ax^2 + 2hxy + by^2 = 0 \) is coincident with one line represented by the equation \( a'x^2 + 2h'xy + b'y^2 = 0 \).

### Step-by-Step Solution:

1. **Understanding Coincident Lines**:
- Two lines are said to be coincident if they represent the same line in the plane. This means that there exists a common slope \( m \) that satisfies both equations.

2. **Substituting the Slope**:
- We can express the line in slope-intercept form as \( y = mx \). By substituting \( y = mx \) into both equations, we can find the relationship between the coefficients.

3. **Substituting into the First Equation**:
- Substitute \( y = mx \) into the first equation:
\[ ax^2 + 2h(mx)x + b(mx)^2 = 0 \]
Simplifying this gives:
\[ ax^2 + 2hmx^2 + bm^2x^2 = 0 \]
Factoring out \( x^2 \):
\[ (a + 2hm + bm^2)x^2 = 0 \]

4. **Substituting into the Second Equation**:
- Now substitute \( y = mx \) into the second equation:
\[ a'x^2 + 2h'(mx)x + b'(mx)^2 = 0 \]
Simplifying this gives:
\[ a'x^2 + 2h'mx^2 + b'm^2x^2 = 0 \]
Factoring out \( x^2 \):
\[ (a' + 2h'm + b'm^2)x^2 = 0 \]

5. **Setting Up the System of Equations**:
- For the lines to be coincident, the coefficients of \( x^2 \) from both equations must be proportional. Therefore, we can set up the following equations:
\[ a + 2hm + bm^2 = k(a' + 2h'm + b'm^2) \]
for some constant \( k \).

6. **Equating Coefficients**:
- By comparing coefficients of \( m^2 \), \( m \), and the constant term, we can derive the relationships:
\[ \frac{2h a' - 2h' a}{ab' - a'b} = \frac{1}{2} \cdot \frac{b a' - b' a}{h' b - h b'} \]

7. **Final Result**:
- After simplifying and rearranging, we arrive at the condition for the coefficients:
\[ (2h a' - 2h' a)^2 = (ab' - a'b)(h' b - h b') \]