1. The problem states we have two parabolic curves derived from birth months and years. 2. For the father: birth month is June (6), birth year ends with 20. 3. For the mother: birth month is August (8), birth year ends with 22. 4. The curves are: $$y_1 = - (6) x^2 + 20 = -6x^2 + 20$$ $$y_2 = 8 x^2 - 22$$ 5. To find intersections, set $y_1 = y_2$: $$-6x^2 + 20 = 8x^2 - 22$$ 6. Bring all terms to one side: $$-6x^2 + 20 - 8x^2 + 22 = 0$$ $$-14x^2 + 42 = 0$$ 7. Simplify: $$-14x^2 = -42$$ $$x^2 = 3$$ 8. Solve for $x$: $$x = \\pm \sqrt{3}$$ So intersections exist at $x = \\pm \sqrt{3}$. 9. Find $y$ values at these $x$ values for $y_1$ or $y_2$: Using $y_1$: $$y = -6(3) + 20 = -18 + 20 = 2$$ Thus intersection points are $(\sqrt{3}, 2)$ and $(-\sqrt{3}, 2)$. 10. Next, find the area between the two curves over the interval $[-\sqrt{3}, \sqrt{3}]$. Area formula: $$A = \int_{-\sqrt{3}}^{\sqrt{3}} |y_1 - y_2| dx$$ Since $y_1 - y_2 = (-6x^2 + 20) - (8x^2 - 22) = -6x^2 + 20 - 8x^2 + 22 = -14x^2 + 42$, and this is positive in the interval (since at $x=0$, value is 42), absolute value simplifies to: $$A = \int_{-\sqrt{3}}^{\sqrt{3}} (42 - 14x^2) dx$$ 11. Evaluate the integral: $$A = \left[42x - \frac{14x^3}{3}\right]_{-\sqrt{3}}^{\sqrt{3}}$$ 12. Calculate at $x=\sqrt{3}$: $$42(\sqrt{3}) - \frac{14(\sqrt{3})^3}{3} = 42\sqrt{3} - \frac{14 \times 3\sqrt{3}}{3} = 42\sqrt{3} - 14\sqrt{3} = 28\sqrt{3}$$ 13. Calculate at $x=-\sqrt{3}$: $$42(-\sqrt{3}) - \frac{14(-\sqrt{3})^3}{3} = -42\sqrt{3} - \frac{14(-3\sqrt{3})}{3} = -42\sqrt{3} + 14\sqrt{3} = -28\sqrt{3}$$ 14. Find difference: $$28\sqrt{3} - (-28\sqrt{3}) = 56\sqrt{3}$$ 15. Therefore, the area between the two curves is: $$\boxed{56\sqrt{3}}$$ This completes the process including intersection and area calculation.