Determine the position of a normal chord of a parabola such that it cuts off of the parabola a segment of minimum area.
1. **Parameterize the Parabola:**
   The given parabola is \( y = ax^2 \). A point on the parabola can be parameterized as \( (p, ap^2) \).

2. **Find the Slope of the Tangent:**
   The slope of the tangent to the parabola at \( (p, ap^2) \) is given by the derivative \( \frac{dy}{dx} = 2ap \).

3. **Find the Slope of the Normal:**
   The slope of the normal to the parabola at \( (p, ap^2) \) is the negative reciprocal of the slope of the tangent:
   \[
   \text{slope of normal} = -\frac{1}{2ap}
   \]

4. **Equation of the Normal:**
   The equation of the normal line at \( (p, ap^2) \) can be written using the point-slope form:
   \[
   y - ap^2 = -\frac{1}{2ap}(x - p)
   \]
   Simplifying, we get:
   \[
   y = -\frac{1}{2ap}x + \left(ap^2 + \frac{p}{2ap}\right)
   \]

5. **Find the Intersection of the Normal with the Parabola:**
   To find the intersection of the normal with the parabola on the opposite side, substitute \( y = ax^2 \) into the equation of the normal:
   \[
   ax^2 = -\frac{1}{2ap}x + \left(ap^2 + \frac{p}{2ap}\right)
   \]
   Rearrange to form a quadratic equation:
   \[
   2a^2px^2 + x - 2a^3p^3 - \frac{1}{2ap} = 0
   \]
   Solving for \( x \), we get:
   \[
   x = -\left(p + \frac{1}{2a^3p}\right)
   \]

6. **Set Up the Integral for the Area:**
   The area of the segment cut off by the chord can be found by integrating the difference between the line and the parabola from \( p \) to \( x \):
   \[
   \text{Area} = \int_p^x \left( -\frac{1}{2ap}t + \left(ap^2 + \frac{p}{2ap}\right) - at^2 \right) dt
   \]
   Substitute \( x = -\left(p + \frac{1}{2a^3p}\right) \) into the integral.

7. **Simplify and Minimize the Area:**
   Simplify the integral and find the expression in terms of \( p \). Then, minimize this expression with respect to \( p \).

8. **Conclusion:**
   After performing the integration and minimization, the position of the normal chord that cuts off the segment of minimum area will be found in terms of \( a \).

The final answer is in terms of \( \boxed{ a } \).