If the line `y=mx+c` is a normal to the parabola `y^2=4ax`, then c is

Options

A. `am(m^2+2)`
B. `am^2(m+2)`
C. `-am(m^2+2)`
D. `-am^2(m+2)`

Answer

To find the value of \( c \) in the equation of the line \( y = mx + c \), which is normal to the parabola given by \( y^2 = 4ax \), we can follow these steps:

### Step 1: Understand the equations
The equation of the parabola is given as:
\[ y^2 = 4ax \]
The equation of the line is given as:
\[ y = mx + c \]

### Step 2: Write the equation of the normal to the parabola
For the parabola \( y^2 = 4ax \), the equation of the normal at a point \( (at^2, 2at) \) on the parabola can be expressed as:
\[ y = -\frac{1}{t}(x - at^2) + 2at \]
This simplifies to:
\[ y = -\frac{1}{t}x + \left(2at + \frac{at^2}{t}\right) = -\frac{1}{t}x + 3at \]

### Step 3: Relate the slope of the normal to the line
The slope of the normal line is \( -\frac{1}{t} \). Since the line \( y = mx + c \) is normal to the parabola, we can equate the slopes:
\[ m = -\frac{1}{t} \]
From this, we can express \( t \) in terms of \( m \):
\[ t = -\frac{1}{m} \]

### Step 4: Substitute \( t \) back to find \( c \)
Now, substituting \( t = -\frac{1}{m} \) into the normal equation:
\[ c = 3a\left(-\frac{1}{m}\right) = -\frac{3a}{m} \]

### Step 5: Final expression for \( c \)
Thus, the value of \( c \) is:
\[ c = -\frac{3a}{m} \]

### Conclusion
The value of \( c \) when the line \( y = mx + c \) is normal to the parabola \( y^2 = 4ax \) is:
\[ c = -\frac{3a}{m} \]