Key Concepts and Formulas

• Equation of a Line: The equation of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
• Area under a curve: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$, is given by $\int_a^b (f(x) - g(x))\, dx$.
• Area of a triangle: The area of a triangle with vertices $(0,0)$, $(x_1, y_1)$, and $(x_2, y_2)$ is $\frac{1}{2} |x_1y_2 - x_2y_1|$.
• AM-GM Inequality: For non-negative numbers $x$ and $y$, $\frac{x+y}{2} \ge \sqrt{xy}$.

Step-by-Step Solution

Step 1: Find the equation of the line PQ. We are given the points $P(a, a^2)$ and $Q(-b, b^2)$. The equation of the line passing through these points is given by: $y - a^2 = \frac{b^2 - a^2}{-b - a}(x - a)$ $y - a^2 = \frac{(b-a)(b+a)}{-(b+a)}(x - a)$ Since $a > 0$ and $b > 0$, $a+b \neq 0$, so we can cancel $a+b$. $y - a^2 = (a-b)(x - a)$ $y = (a-b)x - a^2 + ab + a^2$ $y = (a-b)x + ab$

Step 2: Calculate the area $S_1$ between the line PQ and the parabola $y = x^2$. $S_1 = \int_{-b}^{a} ((a-b)x + ab - x^2) \, dx$ $S_1 = \left[ \frac{(a-b)x^2}{2} + abx - \frac{x^3}{3} \right]_{-b}^{a}$ $S_1 = \left( \frac{(a-b)a^2}{2} + a^2b - \frac{a^3}{3} \right) - \left( \frac{(a-b)b^2}{2} - ab^2 + \frac{b^3}{3}(-1) \right)$ $S_1 = \frac{a^3 - a^2b}{2} + a^2b - \frac{a^3}{3} - \frac{ab^2 - b^3}{2} + ab^2 + \frac{b^3}{3}$ $S_1 = \frac{3a^3 - 3a^2b + 6a^2b - 2a^3 - 3ab^2 + 3b^3 + 6ab^2 + 2b^3}{6}$ $S_1 = \frac{a^3 + 3a^2b + 3ab^2 + b^3}{6}$ $S_1 = \frac{(a+b)^3}{6}$

Step 3: Calculate the area $S_2$ of the triangle OPQ. The vertices are $O(0,0)$, $P(a, a^2)$, and $Q(-b, b^2)$. $S_2 = \frac{1}{2} |a \cdot b^2 - (-b) \cdot a^2|$ $S_2 = \frac{1}{2} |ab^2 + a^2b|$ $S_2 = \frac{1}{2} |ab(a+b)|$ Since $a > 0$ and $b > 0$, $S_2 = \frac{1}{2} ab(a+b)$

Step 4: Calculate the ratio $\frac{S_1}{S_2}$. $\frac{S_1}{S_2} = \frac{\frac{(a+b)^3}{6}}{\frac{ab(a+b)}{2}} = \frac{(a+b)^3}{6} \cdot \frac{2}{ab(a+b)} = \frac{(a+b)^2}{3ab}$ $\frac{S_1}{S_2} = \frac{a^2 + 2ab + b^2}{3ab} = \frac{a^2}{3ab} + \frac{2ab}{3ab} + \frac{b^2}{3ab} = \frac{a}{3b} + \frac{2}{3} + \frac{b}{3a}$ $\frac{S_1}{S_2} = \frac{1}{3} \left( \frac{a}{b} + \frac{b}{a} \right) + \frac{2}{3}$

Step 5: Minimize the ratio using AM-GM. By AM-GM inequality, $\frac{\frac{a}{b} + \frac{b}{a}}{2} \ge \sqrt{\frac{a}{b} \cdot \frac{b}{a}} = \sqrt{1} = 1$ $\frac{a}{b} + \frac{b}{a} \ge 2$ Therefore, $\frac{S_1}{S_2} = \frac{1}{3} \left( \frac{a}{b} + \frac{b}{a} \right) + \frac{2}{3} \ge \frac{1}{3}(2) + \frac{2}{3} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$ The minimum value of $\frac{S_1}{S_2}$ is $\frac{4}{3}$. So, $m = 4$ and $n = 3$. Then $m+n = 4+3 = 7$.

Step 6: Re-evaluating the solution The result is incorrect. Going back to Step 4, we got $\frac{S_1}{S_2} = \frac{a^2 + 2ab + b^2}{3ab} = \frac{1}{3} \left( \frac{a}{b} + 2 + \frac{b}{a} \right)$ Let $x = a/b$. Then $\frac{S_1}{S_2} = \frac{1}{3} \left( x + \frac{1}{x} + 2 \right)$. We know $x + \frac{1}{x} \ge 2$. So $\frac{S_1}{S_2} \ge \frac{1}{3} (2 + 2) = \frac{4}{3}$ when $x=1$ or $a = b$. If $a = b$, then $P = (a, a^2)$ and $Q = (-a, a^2)$. The points become $O(0, 0)$, $P(a, a^2)$, $Q(-a, a^2)$. The minimum value of $\frac{S_1}{S_2}$ is $\frac{4}{3}$, so $m = 4$ and $n = 3$. Then $m+n = 4+3 = 7$.

Step 7: Correcting the Error

There was an error in the problem statement. The correct answer should be 7 and not 2. Let's re-evaluate with the correct answer being 7.

The minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n} = \frac{4}{3}$. Since $\gcd(4, 3) = 1$, we have $m = 4$ and $n = 3$. Thus, $m + n = 4 + 3 = 7$.

Common Mistakes & Tips

• Be careful with signs when evaluating definite integrals.
• Remember the AM-GM inequality and when it applies.
• Double-check algebraic manipulations to avoid errors.
• Memorizing the shortcut for the area of a parabolic segment can save time.

Summary

We found the equation of the line PQ, calculated the areas $S_1$ and $S_2$, and then minimized the ratio $\frac{S_1}{S_2}$ using the AM-GM inequality. The minimum value of the ratio is $\frac{4}{3}$, so $m=4$ and $n=3$, and $m+n=7$.

Final Answer

The final answer is \boxed{7}.