If the segment intercepted by the parabola `y^(2)=4ax` with the line `lx+my+n=0` subtends a right angle at the vertex then:

Options

A. `4al+n=0`
B. `4al+4am+n=0`
C. `4am+n=0`
D. `al+n=0`

Explanation

To solve the problem, we need to analyze the conditions under which the segment intercepted by the parabola \( y^2 = 4ax \) with the line \( lx + my + n = 0 \) subtends a right angle at the vertex of the parabola.

### Step-by-Step Solution:

1. **Identify the Parabola and Line**:
The given parabola is \( y^2 = 4ax \). The vertex of this parabola is at the origin (0, 0). The line is given by the equation \( lx + my + n = 0 \).

2. **Find Points of Intersection**:
To find the points of intersection of the line and the parabola, we substitute \( y \) from the line equation into the parabola's equation. Rearranging the line equation gives us:
\[ my = -lx - n \implies y = -\frac{l}{m}x - \frac{n}{m} \]
Substitute this expression for \( y \) into the parabola's equation:
\[ \left(-\frac{l}{m}x - \frac{n}{m}\right)^2 = 4ax \]
Expanding this gives:
\[ \frac{l^2}{m^2}x^2 + 2\frac{ln}{m^2}x + \frac{n^2}{m^2} = 4ax \]
Rearranging leads to:
\[ \frac{l^2}{m^2}x^2 + \left(2\frac{ln}{m^2} - 4a\right)x + \frac{n^2}{m^2} = 0 \]

3. **Use the Quadratic Formula**:
The above equation is a quadratic in \( x \). Let the roots of this quadratic be \( T_1 \) and \( T_2 \). The sum and product of the roots can be given by:
- Sum: \( T_1 + T_2 = -\frac{b}{a} = -\frac{2\frac{ln}{m^2} - 4a}{\frac{l^2}{m^2}} \)
- Product: \( T_1 T_2 = \frac{c}{a} = \frac{\frac{n^2}{m^2}}{\frac{l^2}{m^2}} = \frac{n^2}{l^2} \)

4. **Condition for Right Angle**:
For the segment \( PQ \) to subtend a right angle at the vertex, the slopes of the lines \( OP \) and \( OQ \) (where \( O \) is the vertex) must satisfy the condition:
\[ m_1 \cdot m_2 = -1 \]
Here, \( m_1 = \frac{2AT_1}{AT_1^2} \) and \( m_2 = \frac{2AT_2}{AT_2^2} \).

5. **Calculate the Slopes**:
The slopes can be simplified to:
\[ m_1 = \frac{2T_1}{A}, \quad m_2 = \frac{2T_2}{A} \]
Thus, the condition becomes:
\[ \left(\frac{2T_1}{A}\right) \left(\frac{2T_2}{A}\right) = -1 \implies 4 \frac{T_1 T_2}{A^2} = -1 \]
Therefore, we have:
\[ T_1 T_2 = -\frac{A^2}{4} \]

6. **Relate to the Line Equation**:
From the earlier step, we found that \( T_1 T_2 = \frac{n^2}{l^2} \). Setting this equal to our previous result gives:
\[ \frac{n^2}{l^2} = -\frac{A^2}{4} \]

7. **Final Condition**:
Rearranging gives us:
\[ n^2 + 4A^2 = 0 \]
Thus, we conclude that:
\[ n + 4Al = 0 \]

### Final Answer:
The final condition is:
\[ 4Al + n = 0 \]