The equation of two straight lines through the point `(x_(1),y_(1))` and perpendicular to the lines given by `ax^(2)+2hxy+by^(2)=0`, is

Options

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

Explanation

To find the equation of two straight lines through the point \((x_1, y_1)\) and perpendicular to the lines given by the equation \(ax^2 + 2hxy + by^2 = 0\), we can follow these steps:

### Step 1: Identify the given equation
The given equation of the conic section is:
\[ ax^2 + 2hxy + by^2 = 0 \]
This represents two straight lines that intersect at the origin.

### Step 2: Determine the slopes of the lines
The slopes of the lines represented by the equation can be found using the formula derived from the quadratic equation:
\[ m^2 + \frac{2h}{a}m + \frac{b}{a} = 0 \]
Let the slopes of the lines be \(m_1\) and \(m_2\).

### Step 3: Find the slopes of the perpendicular lines
The slopes of the lines that are perpendicular to these lines will be the negative reciprocals of \(m_1\) and \(m_2\). If \(m_1\) and \(m_2\) are the slopes of the original lines, then the slopes of the perpendicular lines, say \(m_3\) and \(m_4\), will be:
\[ m_3 = -\frac{1}{m_1}, \quad m_4 = -\frac{1}{m_2} \]

### Step 4: Write the equations of the lines through the point \((x_1, y_1)\)
Using the point-slope form of the equation of a line, the equations of the lines through the point \((x_1, y_1)\) with slopes \(m_3\) and \(m_4\) can be written as:
\[ y - y_1 = m_3(x - x_1) \quad \text{and} \quad y - y_1 = m_4(x - x_1) \]

### Step 5: Substitute the slopes
Substituting \(m_3\) and \(m_4\) into the equations gives us:
\[ y - y_1 = -\frac{1}{m_1}(x - x_1) \quad \text{and} \quad y - y_1 = -\frac{1}{m_2}(x - x_1) \]

### Step 6: Rearranging to standard form
Rearranging these equations to standard form will yield the required equations of the two lines through the point \((x_1, y_1)\).

### Final Result
The final equations of the two lines through the point \((x_1, y_1)\) and perpendicular to the lines given by \(ax^2 + 2hxy + by^2 = 0\) can be expressed as:
\[ b(x - x_1)^2 - 2h(x - x_1)(y - y_1) + a(y - y_1)^2 = 0 \]