JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 21)
Three points $\mathrm{O}(0,0), \mathrm{P}\left(\mathrm{a}, \mathrm{a}^2\right), \mathrm{Q}\left(-\mathrm{b}, \mathrm{b}^2\right), \mathrm{a}>0, \mathrm{~b}>0$, are on the parabola $y=x^2$. Let $\mathrm{S}_1$ be the area of the region bounded by the line $\mathrm{PQ}$ and the parabola, and $\mathrm{S}_2$ be the area of the triangle $\mathrm{OPQ}$. If the minimum value of $\frac{\mathrm{S}_1}{\mathrm{~S}_2}$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to __________.
Answer
7
Explanation
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Equation of $P Q$
$$ \begin{aligned} & y+b=\frac{a^2-b^2}{a+b}\left(x-b^2\right) \\\\ & \Rightarrow y=(a-b) x+a b \end{aligned} $$
$\begin{aligned} & S_1=\int\limits_{-b}^a\left((a-b) x+a b-x^2\right) d x \\\\ & =\left(\frac{(a-b)}{2} x^2+a b x-\frac{x^3}{3}\right)_{-b}^a \\\\ & =\frac{1}{6}(a+b)^3\end{aligned}$
$S_2=\frac{1}{2}\left|\begin{array}{ccc}-b & b^2 & 1 \\ 0 & 0 & 1 \\ a & a^2 & 1\end{array}\right|=\frac{1}{2} a b(a+b)$
$\begin{aligned} \frac{S_1}{S_2} & =\frac{\frac{1}{6}(a+b)^3}{\frac{1}{2} \cdot a b(a+b)} \\\\ & =\frac{1}{3} \frac{(a+b)^2}{a b}=\frac{1}{3}\left(\frac{a}{b}+\frac{b}{a}+2\right)\end{aligned}$
$\begin{aligned} & \because \frac{b}{a}+\frac{a}{b} \geq 2 \\\\ & \Rightarrow\left(\frac{S_1}{S_2}\right)_{\min }=\frac{4}{3}=\frac{m}{n} \\\\ & \Rightarrow m+n=7\end{aligned}$
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