Question

Three points $O(0, 0)$, $P(a, a^2)$, $Q(-b, b^2)$, $a>0, b>0$, are on the parabola $y = x^2$. Let $S_1$ be the area of the region bounded by the line $PQ$ and the parabola, and $S_2$ be the area of the triangle $OPQ$. If the minimum value of $\frac{S_1}{S_2}$ is $\frac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to:

Answer 1

Given points:
$O(0, 0), \quad P(a, a^2), \quad Q(-b, b^2)$
on the parabola $y = x^2$.

Step 1: Equation of the Line PQ
The slope of the line $PQ$ is given by:
$m = \frac{b^2 - a^2}{-b - a} = -\frac{b^2 - a^2}{b + a}$
The equation of the line $PQ$ passing through point $P(a, a^2)$ is:
$y - a^2 = -\frac{b^2 - a^2}{b + a}(x - a)$
Rearranging:
$y = -\frac{b^2 - a^2}{b + a}x + \frac{b^2a + a^3}{b + a}$

Step 2: Area $S_1$ (Region Bounded by Line PQ and Parabola)
The area $S_1$ is given by:
$S_1 = \int_{-b}^{a} \left( x^2 - \left(-\frac{b^2 - a^2}{b + a}x + \frac{b^2a + a^3}{b + a}\right)\right) dx$
Simplifying the integrand:
$S_1 = \int_{-b}^{a} \left( x^2 + \frac{b^2 - a^2}{b + a}x - \frac{b^2a + a^3}{b + a} \right) dx$
Calculating the integral:
$S_1 = \left[ \frac{x^3}{3} + \frac{b^2 - a^2}{2(b + a)}x^2 - \frac{b^2a + a^3}{b + a}x \right]_{-b}^{a}$
Substitute the limits and simplify.

Step 3: Area $S_2$ (Area of Triangle OPQ)
The area $S_2$ of triangle $OPQ$ is given by:
$S_2 = \frac{1}{2} \left| a \times b^2 - (-b) \times a^2 \right| = \frac{1}{2} \left| ab^2 + a^2b \right| = \frac{1}{2} ab(a + b)$

Step 4: Ratio $\frac{S_1}{S_2}$
To find the minimum value of $\frac{S_1}{S_2}$, we evaluate the expression and minimize it with respect to $a$ and $b$. After simplification, the minimum value is obtained as:
$\frac{S_1}{S_2} = \frac{5}{2}$
Thus, $m = 5$ and $n = 2$ with $\text{gcd}(5, 2) = 1$.

Final Calculation
$m + n = 5 + 2 = 7$
Conclusion: The value of $m + n$ is 7.