Find the condition that the slope of one of the lines represented by `ax^2+2hxy+by^2=0` should be n times the slope of the other .

Answer

To find the condition that the slope of one of the lines represented by the equation \( ax^2 + 2hxy + by^2 = 0 \) should be \( n \) times the slope of the other line, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation of the pair of lines:
\[ ax^2 + 2hxy + by^2 = 0 \]
To analyze the slopes, we can divide the entire equation by \( x^2 \):
\[ a + 2h \frac{y}{x} + b \left( \frac{y}{x} \right)^2 = 0 \]
Let \( m = \frac{y}{x} \). Then, we can rewrite the equation as:
\[ bm^2 + 2hm + a = 0 \]

### Step 2: Identify the slopes
This is a quadratic equation in \( m \). The roots of this equation, \( m_1 \) and \( m_2 \), represent the slopes of the two lines. According to Vieta's formulas:
- The sum of the slopes \( m_1 + m_2 = -\frac{2h}{b} \)
- The product of the slopes \( m_1 \cdot m_2 = \frac{a}{b} \)

### Step 3: Set the relationship between slopes
We are given that one slope is \( n \) times the other. Let's assume:
\[ m_2 = n m_1 \]
Now, substituting this into the sum of the slopes:
\[ m_1 + n m_1 = -\frac{2h}{b} \]
This simplifies to:
\[ m_1 (n + 1) = -\frac{2h}{b} \]
Thus, we can express \( m_1 \) as:
\[ m_1 = -\frac{2h}{b(n + 1)} \]

### Step 4: Substitute into the product of slopes
Next, we substitute \( m_2 \) into the product of the slopes:
\[ m_1 \cdot (n m_1) = \frac{a}{b} \]
This gives us:
\[ n m_1^2 = \frac{a}{b} \]
Substituting \( m_1 \) from the earlier equation:
\[ n \left(-\frac{2h}{b(n + 1)}\right)^2 = \frac{a}{b} \]
This leads to:
\[ n \cdot \frac{4h^2}{b^2(n + 1)^2} = \frac{a}{b} \]

### Step 5: Rearranging the equation
Multiplying both sides by \( b^2(n + 1)^2 \) gives:
\[ 4nh^2 = ab(n + 1)^2 \]

### Conclusion
Thus, the condition that the slope of one of the lines is \( n \) times the slope of the other is:
\[ 4nh^2 = ab(n + 1)^2 \]