If three distinct normals are drawn from `(2k, 0)` to the parabola `y^2 = 4x` such that one of them is x-axis and other two are perpendicular, then `k =`

Options

A. 1
B. `1/2`
C. `3/2`
D. none of these

Explanation

To solve the problem, we need to find the value of \( k \) such that three distinct normals can be drawn from the point \( (2k, 0) \) to the parabola \( y^2 = 4x \), where one normal is the x-axis and the other two are perpendicular to each other.

### Step-by-step Solution:

1. **Equation of the Normal to the Parabola**:
The equation of the normal to the parabola \( y^2 = 4x \) at a point \( (at^2, 2at) \) is given by:
\[ y = mx - 2am - am^3 \]
For the parabola \( y^2 = 4x \), we have \( a = 1 \). Thus, the equation simplifies to:
\[ y = mx - 2m - m^3 \]

2. **Substituting the Point**:
We need the normal to pass through the point \( (2k, 0) \). Substituting \( x = 2k \) and \( y = 0 \) into the normal equation gives:
\[ 0 = m(2k) - 2m - m^3 \]
Rearranging this, we have:
\[ m^3 + 2m(1 - k) = 0 \]

3. **Finding the Roots**:
This equation can be factored as:
\[ m(m^2 + 2(1 - k)) = 0 \]
Therefore, one root is \( m = 0 \) and the other roots satisfy:
\[ m^2 + 2(1 - k) = 0 \]
This gives:
\[ m^2 = -2(1 - k) \]

4. **Condition for Perpendicular Normals**:
The two non-zero slopes \( m_1 \) and \( m_2 \) of the normals must be perpendicular. For two slopes to be perpendicular, their product must equal -1:
\[ m_1 m_2 = -1 \]
From the quadratic \( m^2 + 2(1 - k) = 0 \), we can find the product of the roots:
\[ m_1 m_2 = 2(1 - k) \]
Setting this equal to -1 gives:
\[ 2(1 - k) = -1 \]

5. **Solving for \( k \)**:
Now, we solve for \( k \):
\[ 2(1 - k) = -1 \implies 1 - k = -\frac{1}{2} \implies -k = -\frac{1}{2} - 1 \implies -k = -\frac{3}{2} \implies k = \frac{3}{2} \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{\frac{3}{2}} \]