1. **Problem statement:** Find the coordinates of point A on the parabola $y = ax^2$ where an equilateral triangle $OAB$ with side length 4 lies, with $O$ at the origin. 2. **Given:** - Triangle $OAB$ is equilateral with side length 4. - Point $O$ is at $(0,0)$. - Point $A$ lies on the parabola $y = ax^2$. - $OC : 4 = 1 : 2$ so $OC = 2$. - $AC : 4 = \sqrt{3} : 2$ so $AC = 2\sqrt{3}$. 3. **Find coordinates of $A$:** Since $OC$ is the horizontal distance from $O$ to the foot of the perpendicular from $A$ to the x-axis, $C$ is at $(2,0)$. $AC$ is the vertical distance from $C$ to $A$, so $A$ is at $(2, 2\sqrt{3})$. Thus, $A = (2, 2\sqrt{3})$. 4. **Find $a$:** Substitute $A$ into $y = ax^2$: $$2\sqrt{3} = a \cdot 2^2 = 4a$$ Divide both sides by 4: $$\cancel{4}a = \frac{2\sqrt{3}}{\cancel{4}}$$ So, $$a = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$$ 5. **Find area $S$ of equilateral triangle $OAB$:** Area of equilateral triangle with side length $s$ is: $$S = \frac{\sqrt{3}}{4} s^2$$ Substitute $s=4$: $$S = \frac{\sqrt{3}}{4} \times 4^2 = \frac{\sqrt{3}}{4} \times 16 = 4\sqrt{3}$$ **Final answers:** - Coordinates of $A$: $(2, 2\sqrt{3})$ - Value of $a$: $\frac{\sqrt{3}}{2}$ - Area $S$: $4\sqrt{3}$