If `x= my+c` is a normal to the parabola `x^2 = 4ay`, then the value of c is

Options

A. `-2 am - am^3`
B. `2am +am^3`
C. `-(2a)/m - a/m^3`
D. `+(2a)/m + a/m^3`

Answer

To find the value of \( c \) in the equation of the normal to the parabola \( x^2 = 4ay \), we can follow these steps:

### Step 1: Understand the given equations
We have the equation of the parabola:
\[ x^2 = 4ay \]
and the equation of the normal:
\[ x = my + c \]

### Step 2: Differentiate the parabola
To find the slope of the tangent to the parabola at a point \( P(x_1, y_1) \), we differentiate the parabola:
\[ 2x = 4a \frac{dy}{dx} \]
Thus, the slope of the tangent at point \( P \) is:
\[ \frac{dy}{dx} = \frac{x_1}{2a} \]

### Step 3: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, if the slope of the tangent is \( \frac{x_1}{2a} \), then the slope of the normal is:
\[ -\frac{2a}{x_1} \]

### Step 4: Relate the slopes
From the normal equation \( x = my + c \), we can express it in slope-intercept form:
\[ y = \frac{1}{m}x - \frac{c}{m} \]
Thus, the slope of the normal is \( \frac{1}{m} \). Setting the slopes equal gives:
\[ -\frac{2a}{x_1} = \frac{1}{m} \]
From this, we can find \( x_1 \):
\[ x_1 = -2am \]

### Step 5: Substitute \( x_1 \) back into the parabola
Now, substituting \( x_1 \) into the parabola equation to find \( y_1 \):
\[ (-2am)^2 = 4a y_1 \]
This simplifies to:
\[ 4a^2m^2 = 4a y_1 \]
Dividing both sides by \( 4a \) (assuming \( a \neq 0 \)):
\[ y_1 = am^2 \]

### Step 6: Substitute \( y_1 \) into the normal equation
Now substitute \( x_1 \) and \( y_1 \) back into the normal equation:
\[ -2am = m(am^2) + c \]
This simplifies to:
\[ -2am = am^3 + c \]
Rearranging gives:
\[ c = -2am - am^3 \]

### Conclusion
Thus, the value of \( c \) is:
\[ c = -2am - am^3 \]

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