Let there be two prabolas `y^(2)=4x and y^(2)=-8x`. Then the locus of the mid - points of the intercepts between the parabolas made on the lines parallel to the common axis is

Options

A. `y^(2)=16x`
B. `x^(2)=16y`
C. `y^(2)=-8x`
D. `x^(2)=8y`

Answer

To solve the problem, we need to find the locus of the midpoints of the intercepts between the two parabolas \(y^2 = 4x\) and \(y^2 = -8x\) made by lines parallel to the common axis (the x-axis).

### Step-by-Step Solution:

1. **Identify the Parabolas**:
- The first parabola is \(y^2 = 4x\), which opens to the right.
- The second parabola is \(y^2 = -8x\), which opens to the left.

2. **Equation of the Line**:
- We consider a horizontal line parallel to the x-axis, which can be expressed as \(y = h\), where \(h\) is a constant.

3. **Finding Intercepts**:
- For the first parabola \(y^2 = 4x\):
- Substitute \(y = h\) into the equation:
\[ h^2 = 4x \implies x = \frac{h^2}{4} \]
- The point of intersection is \(\left(\frac{h^2}{4}, h\right)\).

- For the second parabola \(y^2 = -8x\):
- Substitute \(y = h\) into the equation:
\[ h^2 = -8x \implies x = -\frac{h^2}{8} \]
- The point of intersection is \(\left(-\frac{h^2}{8}, h\right)\).

4. **Midpoint Calculation**:
- Let the points of intersection be \(A\) and \(B\):
- \(A = \left(\frac{h^2}{4}, h\right)\)
- \(B = \left(-\frac{h^2}{8}, h\right)\)
- The midpoint \(P(\alpha, \beta)\) of \(AB\) is given by:
\[ \alpha = \frac{x_1 + x_2}{2} = \frac{\frac{h^2}{4} - \frac{h^2}{8}}{2} \]
\[ \beta = \frac{y_1 + y_2}{2} = \frac{h + h}{2} = h \]

5. **Simplifying \(\alpha\)**:
- Calculate \(\alpha\):
\[ \alpha = \frac{\frac{h^2}{4} - \frac{h^2}{8}}{2} = \frac{\frac{2h^2}{8} - \frac{h^2}{8}}{2} = \frac{\frac{h^2}{8}}{2} = \frac{h^2}{16} \]

6. **Relating \(\alpha\) and \(\beta\)**:
- We have:
\[ \alpha = \frac{h^2}{16}, \quad \beta = h \]
- From \(\beta = h\), we can express \(h\) in terms of \(\beta\):
\[ h = \beta \]
- Substitute \(h\) back into the equation for \(\alpha\):
\[ \alpha = \frac{\beta^2}{16} \]

7. **Finding the Locus**:
- The relationship between \(\alpha\) and \(\beta\) can be rewritten as:
\[ x = \frac{y^2}{16} \]
- Rearranging gives:
\[ y^2 = 16x \]

### Conclusion:
The locus of the midpoints of the intercepts between the two parabolas made by lines parallel to the common axis is given by the equation:
\[ y^2 = 16x \]