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. `klt1`
B. `kgt1`
C. `kle1`
D. `kge1`

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 \( y^2 = 4x \). One of these normals is the x-axis, and the other two are perpendicular to each other.

### Step 1: Write the equation of the normal to the parabola
The equation of the normal to the parabola \( y^2 = 4x \) at a point where the slope of the normal is \( m \) is given by:
\[ y = mx - 2m - m^3 \]
Here, \( a = 1 \) for the parabola \( y^2 = 4x \).

### Step 2: Substitute the point \( (2k, 0) \) into the normal equation
Since the normal passes through the point \( (2k, 0) \), we substitute \( x = 2k \) and \( y = 0 \) into the normal equation:
\[ 0 = m(2k) - 2m - m^3 \]
This simplifies to:
\[ m(2k - 2) = m^3 \]
Rearranging gives us:
\[ m^3 - 2m + 2m = 0 \implies m^3 + 2m - 2k = 0 \]

### Step 3: Factor the cubic equation
We can factor this cubic equation. We know one root is \( m = 0 \) (which corresponds to the x-axis). Thus, we can factor \( m \) out:
\[ m(m^2 + 2) - 2k = 0 \]
This implies:
\[ m^3 + 2m - 2k = 0 \]

### Step 4: Find the other roots
The remaining roots of the cubic equation can be found using the quadratic formula. The quadratic part is:
\[ m^2 + 2 = 2k \]
This gives:
\[ m^2 = 2k - 2 \]
The roots will be real if:
\[ 2k - 2 \geq 0 \implies k \geq 1 \]

### Step 5: Ensure the two normals are perpendicular
For the two normals to be perpendicular, the slopes \( m_1 \) and \( m_2 \) must satisfy:
\[ m_1 \cdot m_2 = -1 \]
From the previous step, we have:
\[ m_1 = \sqrt{2k - 2}, \quad m_2 = -\sqrt{2k - 2} \]
Thus:
\[ \sqrt{2k - 2} \cdot (-\sqrt{2k - 2}) = - (2k - 2) = -1 \]
This gives:
\[ 2k - 2 = 1 \implies 2k = 3 \implies k = \frac{3}{2} \]

### Conclusion
The value of \( k \) is:
\[ \boxed{\frac{3}{2}} \]