The product of the perpendicular drawn from the point `(x_(1),y_(1))` on the lines represented by `ax^(2)+2hxy+by^(2)=0` is ………

Answer

To find the product of the perpendiculars drawn from the point \((x_1, y_1)\) to the lines represented by the equation \(ax^2 + 2hxy + by^2 = 0\), we can follow these steps:

### Step 1: Identify the Lines
The given equation \(ax^2 + 2hxy + by^2 = 0\) represents two straight lines. We can express these lines in slope-intercept form as:
\[ y = m_1x \quad \text{and} \quad y = m_2x \]
where \(m_1\) and \(m_2\) are the slopes of the lines.

### Step 2: Use the Relationships of Roots
From the quadratic equation formed by the slopes, we know:
- The sum of the roots (slopes) is given by:
\[ m_1 + m_2 = -\frac{2h}{b} \]
- The product of the roots (slopes) is given by:
\[ m_1 m_2 = \frac{a}{b} \]

### Step 3: Calculate the Perpendicular Distances
The perpendicular distance \(d\) from the point \((x_1, y_1)\) to the line \(y = mx\) is given by the formula:
\[ d = \frac{|y_1 - mx_1|}{\sqrt{1 + m^2}} \]
For both lines, the distances will be:
\[ d_1 = \frac{|y_1 - m_1x_1|}{\sqrt{1 + m_1^2}} \quad \text{and} \quad d_2 = \frac{|y_1 - m_2x_1|}{\sqrt{1 + m_2^2}} \]

### Step 4: Find the Product of the Distances
The product of the perpendicular distances from the point \((x_1, y_1)\) to the two lines is:
\[ P = d_1 \cdot d_2 = \left(\frac{|y_1 - m_1x_1|}{\sqrt{1 + m_1^2}}\right) \cdot \left(\frac{|y_1 - m_2x_1|}{\sqrt{1 + m_2^2}}\right) \]

### Step 5: Substitute the Values
Substituting the expressions for \(m_1\) and \(m_2\):
\[ P = \frac{|y_1 - m_1x_1| \cdot |y_1 - m_2x_1|}{\sqrt{(1 + m_1^2)(1 + m_2^2)}} \]

### Step 6: Simplify the Expression
Using the relationships \(m_1 + m_2 = -\frac{2h}{b}\) and \(m_1 m_2 = \frac{a}{b}\), we can simplify the numerator and denominator:
1. The numerator becomes:
\[ |y_1^2 - (m_1 + m_2)x_1y_1 + m_1m_2x_1^2| = |y_1^2 + \frac{2h}{b}x_1y_1 + \frac{a}{b}x_1^2| \]
2. The denominator simplifies to:
\[ \sqrt{1 + m_1^2 + m_2^2 + m_1^2m_2^2} \]

### Final Result
Thus, the product of the perpendiculars drawn from the point \((x_1, y_1)\) to the lines represented by the equation \(ax^2 + 2hxy + by^2 = 0\) is:
\[ P = \frac{|b(y_1^2 + \frac{2h}{b}x_1y_1 + \frac{a}{b}x_1^2)|}{\sqrt{(a - b)^2 + 4h^2}} \]