Let $ a,\ b$ are positive constants. Determin the value of a positive number $ m$ such that the areas of four parts of the region bounded by two parabolas $ y\equal{}ax^2\minus{}b,\ y\equal{}\minus{}ax^2\plus{}b$ and the line $ y\equal{}mx$ have equal area.
1. **Identify the points of intersection:**
   The parabolas \( y = ax^2 - b \) and \( y = -ax^2 + b \) intersect the line \( y = mx \) at points \( C \) and \( D \). We need to find these points of intersection.

2. **Solve for \( k \) and \( h \):**
   - For \( y = mx \) and \( y = -ax^2 + b \):
     \[
     mx = -ax^2 + b \implies ax^2 + mx - b = 0
     \]
     Solving this quadratic equation for \( x \):
     \[
     x = \frac{-m \pm \sqrt{m^2 + 4ab}}{2a}
     \]
     Let \( k = \frac{-m + \sqrt{m^2 + 4ab}}{2a} \) and \( -k = \frac{-m - \sqrt{m^2 + 4ab}}{2a} \).

   - For \( y = mx \) and \( y = ax^2 - b \):
     \[
     mx = ax^2 - b \implies ax^2 - mx - b = 0
     \]
     Solving this quadratic equation for \( x \):
     \[
     x = \frac{m \pm \sqrt{m^2 + 4ab}}{2a}
     \]
     Let \( h = \frac{m + \sqrt{m^2 + 4ab}}{2a} \) and \( -h = \frac{m - \sqrt{m^2 + 4ab}}{2a} \).

3. **Calculate the areas:**
   - The area between the parabolas \( y = ax^2 - b \) and \( y = -ax^2 + b \) from \( x = -\sqrt{b/a} \) to \( x = \sqrt{b/a} \):
     \[
     \text{Area} = 2 \int_{0}^{\sqrt{b/a}} (b - ax^2) \, dx = 2 \left[ bx - \frac{a}{3}x^3 \right]_{0}^{\sqrt{b/a}} = \frac{4b\sqrt{b/a}}{3}
     \]

   - The area between the line \( y = mx \) and the parabola \( y = -ax^2 + b \) from \( x = -k \) to \( x = k \):
     \[
     \text{Area} = 2 \int_{0}^{k} (b - ax^2 - mx) \, dx
     \]

   - The area between the line \( y = mx \) and the parabola \( y = ax^2 - b \) from \( x = -h \) to \( x = h \):
     \[
     \text{Area} = 2 \int_{0}^{h} (mx - ax^2 + b) \, dx
     \]

4. **Equate the areas:**
   Since the areas of the four regions are equal, we have:
   \[
   \frac{4b\sqrt{b/a}}{3} = 2 \int_{0}^{k} (b - ax^2 - mx) \, dx + 2 \int_{0}^{h} (mx - ax^2 + b) \, dx
   \]

5. **Solve for \( m \):**
   By solving the above equation, we find:
   \[
   m = 2 \sqrt{(2^{2/3} - 1) ab}
   \]

The final answer is \( \boxed{ m = 2 \sqrt{(2^{2/3} - 1) ab} } \).