The line `lx+my+n=0` is a normal to the parabola `y^2 = 4ax` if

Options

A. `al(l^(2)+2m^(2))+m^(2)n=0`
B. `al(l^(2)+2m^(2))+m^(2)n`
C. `al(2l^(2)+2m^(2))+m^(2)n=0`
D. `al(2l^(2)+2m^(2))+m^(2)n`

Explanation

To determine the condition under which the line \( lx + my + n = 0 \) is normal to the parabola \( y^2 = 4ax \), we can follow these steps:

### Step-by-Step Solution:

1. **Convert the Line Equation**:
The given line equation is \( lx + my + n = 0 \). We can rearrange this into the slope-intercept form \( y = mx + c \).
\[ my = -lx - n \implies y = -\frac{l}{m}x - \frac{n}{m} \]
Here, the slope \( m' = -\frac{l}{m} \) and the y-intercept \( c = -\frac{n}{m} \).

2. **Identify the Condition for Normality**:
For the line to be normal to the parabola \( y^2 = 4ax \), it must satisfy the condition:
\[ c = -2am - am^3 \]
where \( m \) is the slope of the line.

3. **Substituting for \( c \)**:
From our earlier step, we have \( c = -\frac{n}{m} \). Therefore, we set:
\[ -\frac{n}{m} = -2a - a\left(-\frac{l}{m}\right)^3 \]

4. **Simplifying the Equation**:
Substitute \( m' = -\frac{l}{m} \):
\[ -\frac{n}{m} = -2a + \frac{al^3}{m^3} \]
Multiplying through by \( -m \):
\[ n = 2am + \frac{al^3}{m^2} \]

5. **Rearranging**:
Rearranging gives us:
\[ 2am + \frac{al^3}{m^2} - n = 0 \]

6. **Finding a Common Denominator**:
To combine the terms, we can multiply through by \( m^2 \):
\[ 2am^3 + al^3 - nm^2 = 0 \]

7. **Final Condition**:
This leads us to the final condition that must be satisfied for the line to be normal to the parabola:
\[ 2a m^3 + al^3 + nm^2 = 0 \]