If the parabols `y^(2) = 4kx (k gt 0)` and `y^(2) = 4 (x-1)` do not have a common normal other than the axis of parabola, then `k in`

Options

A. `(0,1)`
B. `(2,oo)`
C. `(3,oo)`
D. `(0,oo)`

Explanation

To solve the problem, we need to find the values of \( k \) such that the parabolas \( y^2 = 4kx \) and \( y^2 = 4(x - 1) \) do not have a common normal other than the axis of the parabola.

### Step-by-Step Solution:

1. **Identify the equations of the parabolas:**
- The first parabola is \( y^2 = 4kx \).
- The second parabola is \( y^2 = 4(x - 1) \).

2. **Write the equation of the normal for the first parabola:**
- For the parabola \( y^2 = 4kx \), the equation of the normal at a point with slope \( m \) is given by:
\[ y = mx - 2km - km^3 \]

3. **Write the equation of the normal for the second parabola:**
- For the parabola \( y^2 = 4(x - 1) \), the equation of the normal is:
\[ y = mx - 1 - 2m - m^3 \]
- This simplifies to:
\[ y = mx - 3m - m^3 \]

4. **Set the two normal equations equal to each other:**
- Since we want to find the common normal, we equate the two equations:
\[ mx - 2km - km^3 = mx - 3m - m^3 \]
- Cancel \( mx \) from both sides:
\[ -2km - km^3 = -3m - m^3 \]

5. **Rearranging the equation:**
- Rearranging gives:
\[ (k - 1)m^3 + (2k - 3)m = 0 \]
- Factor out \( m \):
\[ m \left( (k - 1)m^2 + (2k - 3) \right) = 0 \]

6. **Finding the values of \( m \):**
- This gives us two cases:
- \( m = 0 \) (the slope of the normal parallel to the x-axis).
- The quadratic equation \( (k - 1)m^2 + (2k - 3) = 0 \).

7. **Condition for no common normal other than the axis:**
- For \( m = 0 \) to be the only solution, the quadratic must have no real solutions. This occurs when the discriminant is less than zero:
\[ (2k - 3)^2 - 4(k - 1)(0) < 0 \]
- Simplifying gives:
\[ (2k - 3)^2 < 0 \]
- This is always true, so we need to analyze the quadratic itself.

8. **Set the quadratic to be less than zero:**
- The quadratic \( (k - 1)m^2 + (2k - 3) < 0 \) must hold for all \( m \):
- This is true if \( k - 1 < 0 \) (i.e., \( k < 1 \)) and \( 2k - 3 < 0 \) (i.e., \( k < \frac{3}{2} \)).
- Therefore, we have \( k < 1 \) and \( k < \frac{3}{2} \).

9. **Combine the conditions:**
- Since \( k > 0 \) is also given, we combine the conditions:
- \( 0 < k < 1 \) or \( k > \frac{3}{2} \).

10. **Final answer:**
- Thus, the region where \( k \) belongs is:
\[ k \in (0, 1) \cup \left(\frac{3}{2}, \infty\right) \]