The locus of the midpoints of the portion of the normal to the parabola `y^(2)=16x` intercepted between the curve and the axis is another parabola whose latus rectum is ___________ .

Explanation

To solve the problem, we need to find the locus of the midpoints of the portion of the normal to the parabola \( y^2 = 16x \) that is intercepted between the curve and the x-axis. The parabola can be compared to the standard form \( y^2 = 4ax \), where \( a = 4 \).

### Step-by-Step Solution:

1. **Identify the Parabola**:
The given parabola is \( y^2 = 16x \). Here, comparing it with the standard form \( y^2 = 4ax \), we find that \( a = 4 \).

2. **Find the Point on the Parabola**:
Let the point on the parabola be \( P(k^2, 2k) \), where \( k \) is a parameter.

3. **Determine the Slope of the Tangent**:
The slope of the tangent to the parabola at point \( P \) can be derived from the derivative:
\[ \frac{dy}{dx} = \frac{2a}{y} = \frac{8}{2k} = \frac{4}{k} \]

4. **Find the Slope of the Normal**:
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{y}{2a} = -\frac{2k}{8} = -\frac{k}{4} \]

5. **Equation of the Normal**:
The equation of the normal line at point \( P \) can be expressed as:
\[ y - 2k = -\frac{k}{4}(x - k^2) \]

6. **Intercepts of the Normal**:
To find the intercepts, we set \( y = 0 \) to find the x-intercept:
\[ 0 - 2k = -\frac{k}{4}(x - k^2) \]
Rearranging gives:
\[ x = k^2 + \frac{8}{k} \]

7. **Midpoint Calculation**:
The x-intercept is \( \left(k^2 + \frac{8}{k}, 0\right) \) and the point \( P \) is \( (k^2, 2k) \). The midpoint \( M \) of the segment \( PQ \) is given by:
\[ M\left(\frac{k^2 + \left(k^2 + \frac{8}{k}\right)}{2}, \frac{2k + 0}{2}\right) = \left(k^2 + \frac{4}{k}, k\right) \]

8. **Locus of Midpoints**:
Let \( M(h, k) \) where \( h = k^2 + \frac{4}{k} \). To eliminate \( k \), we can express \( k \) in terms of \( h \):
\[ k^3 - hk + 4 = 0 \]
This represents a cubic equation in \( k \).

9. **Finding the Locus**:
By substituting \( k = \frac{y}{2} \) into the equation, we can derive the equation of the locus:
\[ y^2 = 4\left(x - 4\right) \]
This is a parabola of the form \( y^2 = 4a(x - h) \) where \( a = 4 \).

10. **Latus Rectum Calculation**:
The latus rectum of a parabola \( y^2 = 4ax \) is given by \( 4a \). Here, since \( a = 4 \):
\[ \text{Latus Rectum} = 4 \times 4 = 16 \]

### Final Answer:
The latus rectum of the parabola whose locus we found is **16**.