Three normals are drawn to the parabola `y^(2) = 4x` from the point (c,0). These normals are real and distinct when

Options

A. c=0
B. c=1
C. c=2
D. c=3

Answer

To solve the problem, we need to determine the conditions under which three normals drawn from the point \( (c, 0) \) to the parabola \( y^2 = 4x \) are real and distinct.

### Step-by-Step Solution:

1. **Identify the equation of the parabola**:
The given parabola is \( y^2 = 4x \). This is a standard form of a parabola where \( a = 1 \).

2. **Equation of the normal to the parabola**:
The equation of the normal to the parabola at a point where the slope is \( m \) is given by:
\[ y = mx - 2am - am^3 \]
For our parabola, substituting \( a = 1 \):
\[ y = mx - 2m - m^3 \]

3. **Substituting the point \( (c, 0) \)**:
Since the point \( (c, 0) \) lies on the normal, we substitute \( y = 0 \) and \( x = c \):
\[ 0 = mc - 2m - m^3 \]
Rearranging gives:
\[ m^3 + 2m - mc = 0 \]

4. **Factoring the equation**:
We can factor out \( m \):
\[ m(m^2 + 2 - c) = 0 \]
This gives us one solution \( m = 0 \) and the quadratic:
\[ m^2 + 2 - c = 0 \]

5. **Finding conditions for distinct normals**:
For the normals to be real and distinct, the quadratic \( m^2 + 2 - c = 0 \) must have two distinct real solutions. This occurs when the discriminant is positive:
\[ \Delta = b^2 - 4ac = 0^2 - 4(1)(2 - c) > 0 \]
Simplifying this gives:
\[ -4(2 - c) > 0 \implies 2 - c < 0 \implies c > 2 \]

6. **Conclusion**:
Therefore, the condition for three normals to be real and distinct is:
\[ c > 2 \]

### Final Answer:
The normals are real and distinct when \( c > 2 \).