If the line `y = mx` bisects the angle between the lines `ax^2 + 2h xy + by^2 = 0` then ` m ` is a root of the quadratic equation

Options

A. `h(m^(2)-1)+m(b-1)=0`
B. `h(m^(2)-1)+m(a-b)`
C. `h(m^(2)+1)+m(a-b)=0`
D. none of these

Explanation

To solve the problem, we need to determine the relationship between the slope \( m \) of the line \( y = mx \) and the coefficients \( a, b, h \) of the conic section represented by the equation \( ax^2 + 2hxy + by^2 = 0 \). The line \( y = mx \) bisects the angle between the two lines represented by the conic.

### Step-by-Step Solution:

1. **Identify the given equation**:
The equation of the pair of straight lines is given by:
\[ ax^2 + 2hxy + by^2 = 0 \]

2. **Use the angle bisector condition**:
The equation of the angle bisectors of the lines represented by the conic is given by:
\[ \frac{x^2 - y^2}{xy} = \frac{a - b}{h} \]

3. **Cross-multiply**:
Cross-multiplying gives:
\[ h(x^2 - y^2) = (a - b)xy \]

4. **Rearranging the equation**:
Rearranging the equation, we have:
\[ a h x^2 - (a - b)xy - h y^2 = 0 \]

5. **Substitute \( y = mx \)**:
Since \( y = mx \), substitute this into the equation:
\[ a h x^2 - (a - b) x(mx) - h (mx)^2 = 0 \]
This simplifies to:
\[ a h x^2 - (a - b) mx^2 - h m^2 x^2 = 0 \]

6. **Factor out \( x^2 \)**:
Factoring out \( x^2 \) (assuming \( x \neq 0 \)):
\[ (ah - (a - b)m - hm^2)x^2 = 0 \]

7. **Set the quadratic equation**:
For the equation to hold true, the expression inside the parentheses must equal zero:
\[ ah - (a - b)m - hm^2 = 0 \]

8. **Rearranging to standard quadratic form**:
Rearranging gives us:
\[ hm^2 + (a - b)m - ah = 0 \]

9. **Conclusion**:
This is a quadratic equation in \( m \) which shows that \( m \) is a root of the quadratic equation:
\[ hm^2 + (a - b)m - ah = 0 \]