1. **State the problem:** We have a parabola defined by the equation $y = mx^2$ and two points on the curve: $(4, A)$ and $(8, B)$. The derivative is given by $\frac{dy}{dx} = 2mx$. The vertical distance between points A and B on the y-axis is 12 units. 2. **Write the expressions for A and B:** Since $y = mx^2$, at $x=4$, $A = m \times 4^2 = 16m$. At $x=8$, $B = m \times 8^2 = 64m$. 3. **Use the distance between A and B:** The vertical distance between points A and B is $|B - A| = 12$. Substitute the values: $$|64m - 16m| = 12$$ $$|48m| = 12$$ 4. **Solve for $m$:** $$48|m| = 12$$ $$|m| = \frac{12}{48} = \frac{1}{4}$$ So, $m = \pm \frac{1}{4}$. 5. **Interpretation:** The parabola can open upwards or downwards depending on the sign of $m$. Both satisfy the distance condition. **Final answer:** $$m = \pm \frac{1}{4}$$