The condition that the line lx +my+n =0 to be a normal to `y^(2)=4ax` is

Options

A. `al^(2)+2alm^(2)+m^(2)n =0`
B. `al^(3) +2alm^(2)+m^(2)n=0`
C. `al^(2)+2alm^(2)+m^(2)n^(2)=0`
D. `al^(3)+alm^(2)+m^(2) n=0`

Answer

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

### Step 1: Understand the Parabola and the Line
The given parabola is \( y^2 = 4ax \), which opens to the right. The equation of the line is given as \( lx + my + n = 0 \).

### Step 2: Rearrange the Line Equation
We can rearrange the line equation to express \( y \) in terms of \( x \):
\[ my = -lx - n \implies y = -\frac{l}{m}x - \frac{n}{m} \]

### Step 3: Identify the Slope of the Line
The slope \( m \) of the line is \( -\frac{l}{m} \).

### Step 4: Condition for Normality
For a line to be a normal to a parabola, the slope of the normal line at a point on the parabola must be equal to the slope of the line. The slope of the tangent to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ \text{slope of tangent} = \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t} \]
Thus, the slope of the normal at this point is:
\[ \text{slope of normal} = -t \]

### Step 5: Set the Slopes Equal
Setting the slope of the normal equal to the slope of the line:
\[ -t = -\frac{l}{m} \implies t = \frac{l}{m} \]

### Step 6: Substitute \( t \) into the Normal Condition
The equation of the normal to the parabola at the point \( (at^2, 2at) \) is:
\[ y - 2at = -t(x - at^2) \]
Substituting \( t = \frac{l}{m} \):
\[ y - 2a\left(\frac{l}{m}\right) = -\frac{l}{m}\left(x - a\left(\frac{l}{m}\right)^2\right) \]

### Step 7: Rearranging the Normal Equation
Rearranging gives us:
\[ y = -\frac{l}{m}x + 2a\left(\frac{l}{m}\right) + a\left(\frac{l^2}{m^2}\right) \]

### Step 8: Compare with the Line Equation
Now we want to compare this with the line equation \( y = -\frac{l}{m}x - \frac{n}{m} \). Thus, we have:
\[ -\frac{n}{m} = 2a\left(\frac{l}{m}\right) + a\left(\frac{l^2}{m^2}\right) \]

### Step 9: Multiply through by \( m \)
Multiplying through by \( m \) gives:
\[ -n = 2al + \frac{al^2}{m} \]

### Step 10: Rearranging to Find the Condition
Rearranging gives us:
\[ al^2 + 2alm + mn = 0 \]

### Conclusion
Thus, the condition that the line \( lx + my + n = 0 \) is a normal to the parabola \( y^2 = 4ax \) is:
\[ al^2 + 2alm + mn = 0 \]