Area Under the Curve
Area Ratio — Parabola and Triangle
nta_pyq_2024_jan
Grade 12
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:
Step-by-Step Solution
Key Concept: Using the standard result that area between a chord and a parabola $y=x^2$ is $\frac{1}{6}|a+b|^3$ (for endpoints at $x=a$ and $x=-b$), and triangle area via determinant formula. Minimize the ratio using AM-GM.
$S_2=\frac{1}{2}|ab^2+a^2b|=\frac{ab(a+b)}{2}$. $S_1=\int_{-b}^{a}[(a-b)x+ab-x^2]dx=\frac{(a+b)^3}{6}$. $\frac{S_1}{S_2}=\frac{(a+b)^2}{3ab}=\frac{1}{3}[\frac{a}{b}+\frac{b}{a}+2]\ge\frac{4}{3}$. So $m=4$, $n=3$, $m+n=7$.
Correct Answer: 7