If y=mx+4 is common tangent to parabolas `y^(2)=4x and x^(2)=2by`. Then value of b is

Options

A. (a)128
B. (b)`-128`
C. (c)`-64`
D. (d)`-32`

Answer

To solve the problem, we need to find the value of \( b \) such that the line \( y = mx + 4 \) is a common tangent to the parabolas \( y^2 = 4x \) and \( x^2 = 2by \).

### Step-by-Step Solution:

1. **Identify the first parabola**:
The first parabola is given by the equation \( y^2 = 4x \). This can be compared to the standard form \( y^2 = 4ax \), where \( a = 1 \).

2. **Equation of the tangent line**:
The general equation of the tangent to the parabola \( y^2 = 4ax \) is given by:
\[ y = mx + \frac{a}{m} \]
For our parabola, substituting \( a = 1 \):
\[ y = mx + \frac{1}{m} \]

3. **Set the tangent equal to the given line**:
We know the line is \( y = mx + 4 \). By comparing the two equations:
\[ \frac{1}{m} = 4 \]
Solving for \( m \):
\[ m = \frac{1}{4} \]

4. **Identify the second parabola**:
The second parabola is given by \( x^2 = 2by \). We can rearrange this to express \( y \):
\[ y = \frac{x^2}{2b} \]

5. **Substitute the tangent line into the second parabola**:
Substitute \( y = mx + 4 \) into the equation of the second parabola:
\[ mx + 4 = \frac{x^2}{2b} \]
Rearranging gives:
\[ x^2 - 2b(mx + 4) = 0 \]
Expanding this:
\[ x^2 - 2bmx - 8b = 0 \]

6. **Identify coefficients for the quadratic**:
The quadratic equation is in the form \( Ax^2 + Bx + C = 0 \):
- \( A = 1 \)
- \( B = -2bm \)
- \( C = -8b \)

7. **Condition for tangency**:
For the line to be a tangent to the parabola, the discriminant must be zero:
\[ D = B^2 - 4AC = 0 \]
Plugging in the coefficients:
\[ (-2bm)^2 - 4(1)(-8b) = 0 \]
This simplifies to:
\[ 4b^2m^2 + 32b = 0 \]

8. **Factor the equation**:
Factoring out \( 4b \):
\[ 4b(bm^2 + 8) = 0 \]
Thus, we have two cases:
- \( b = 0 \)
- \( bm^2 + 8 = 0 \)

9. **Solve for \( b \)**:
If \( b = 0 \), the second parabola degenerates. So we solve the second case:
\[ bm^2 + 8 = 0 \implies b = -\frac{8}{m^2} \]
Substituting \( m = \frac{1}{4} \):
\[ b = -\frac{8}{\left(\frac{1}{4}\right)^2} = -\frac{8}{\frac{1}{16}} = -8 \times 16 = -128 \]

### Final Answer:
Thus, the value of \( b \) is \( \boxed{-128} \).