The pole of the line lx+my+n=0 with respect to the parabola `y^(2)=4ax,` is

Options

A. `(n/l,-(2am)/l)`
B. `(n/m,-(2am)/m)`
C. `(n/m,-(2al)/m)`
D. none of these

Explanation

To find the pole of the line \( lx + my + n = 0 \) with respect to the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Understand the concept of the pole
The pole of a line with respect to a conic section (in this case, a parabola) is a point from which the tangents drawn to the parabola correspond to the given line. The equation of the polar of a point \( (x_1, y_1) \) with respect to the parabola \( y^2 = 4ax \) is given by:
\[ yy_1 = 2ax + x_1 \]

### Step 2: Set up the equation of the polar
We can rewrite the equation of the polar as:
\[ 2ax - yy_1 + x_1 = 0 \]

### Step 3: Equate the polar and the line
Since the line \( lx + my + n = 0 \) is the polar of the point \( (x_1, y_1) \), we can equate the coefficients of \( x \), \( y \), and the constant term from both equations:
1. Coefficient of \( x \): \( l = 2a \)
2. Coefficient of \( y \): \( m = -y_1 \)
3. Constant term: \( n = x_1 \)

### Step 4: Solve for \( x_1 \) and \( y_1 \)
From the equations we derived:
- From \( n = x_1 \), we have:
\[ x_1 = n \]
- From \( m = -y_1 \), we have:
\[ y_1 = -\frac{m}{l} \]

### Step 5: Substitute \( l \) from the first equation
From \( l = 2a \), we can substitute \( l \) into the equation for \( y_1 \):
\[ y_1 = -\frac{m}{2a} \]

### Step 6: Write the final coordinates of the pole
Thus, the coordinates of the pole \( (x_1, y_1) \) are:
\[ (x_1, y_1) = \left( n, -\frac{m}{2a} \right) \]

### Final Answer
The pole of the line \( lx + my + n = 0 \) with respect to the parabola \( y^2 = 4ax \) is:
\[ \left( n, -\frac{m}{2a} \right) \]

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