Three points O(0,0),P(a, a^2),and Q(-b, b^2) | vedclass

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(C) The area of triangle $OPQ$ is given by $S_2 = \frac{1}{2} |x_O(y_P - y_Q) + x_P(y_Q - y_O) + x_Q(y_O - y_P)| = \frac{1}{2} |0 + a(b^2 - 0) + (-b)(

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$7$

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(C) The area of triangle $OPQ$ is given by $S_2 = \frac{1}{2} |x_O(y_P - y_Q) + x_P(y_Q - y_O) + x_Q(y_O - y_P)| = \frac{1}{2} |0 + a(b^2 - 0) + (-b)(0 - a^2)| = \frac{1}{2} (ab^2 + a^2b) = \frac{ab(a+b)}{2}$.The equation of line $PQ$ passing through $(a, a^2)$ and $(-b, b^2)$ is $y - a^2 = \frac{b^2 - a^2}{-b - a}(x - a)$,which simplifies to $y - a^2 = (a - b)(x - a)$,so $y = (a - b)x + ab$.The area $S_1$ bounded by the parabola $y = x^2$ and the line $y = (a - b)x + ab$ is $S_1 = \int_{-b}^{a} ((a - b)x + ab - x^2) dx = \left[ (a - b)\frac{x^2}{2} + abx - \frac{x^3}{3} \right]_{-b}^{a}$.Evaluating the integral: $S_1 = \left( (a - b)\frac{a^2}{2} + a^2b - \frac{a^3}{3} \right) - \left( (a - b)\frac{b^2}{2} - ab^2 + \frac{b^3}{3} \right) = \frac{a^3 - a^2b + 2a^2b - 2a^3/3 - ab^2 + b^3/2 + ab^2 - b^3/3}{1} = \frac{a^3}{6} + \frac{a^2b}{2} + \frac{ab^2}{2} + \frac{b^3}{6} = \frac{(a+b)^3}{6}$.Now,$\frac{S_1}{S_2} = \frac{(a+b)^3 / 6}{ab(a+b) / 2} = \frac{(a+b)^2}{3ab} = \frac{1}{3} \left( \frac{a^2 + b^2 + 2ab}{ab} \right) = \frac{1}{3} \left( \frac{a}{b} + \frac{b}{a} + 2 \right)$.Since $a, b > 0$,by $AM$-$GM$ inequality,$\frac{a}{b} + \frac{b}{a} \ge 2$. Thus,the minimum value is $\frac{1}{3}(2 + 2) = \frac{4}{3}$.Given $\frac{m}{n} = \frac{4}{3}$,we have $m = 4$ and $n = 3$. Therefore,$m + n = 4 + 3 = 7$.

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