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

Correct Answer: 7

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.