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QJEEM24Maca41: JEE Main 2024 Mathematics - Class 12 JEEM24Maca41 Solution: Three points , are on the parabola . Let be the area of the…

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=x2y=x^2 . Let S1\mathrm{S}_1 be the area of the region bounded by the line PQ\mathrm{PQ} and the parabola, and S2\mathrm{S}_2 be the area of the triangle OPQ\mathrm{OPQ} . If the minimum value of S1 S2\frac{\mathrm{S}_1}{\mathrm{~S}_2} is mn,gcd⁡(m,n)=1\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1 , then m+n\mathrm{m}+\mathrm{n} is equal to __________.

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JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Area Under The Curves Question 32 English Explanation

Equation of PQP Q

amp;y+b=a2−b2a+b(x−b2)amp;⇒y=(a−b)x+ab\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}

amp;S1=∫−ba((a−b)x+ab−x2)dxamp;=((a−b)2x2+abx−x33)−baamp;=16(a+b)3\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}

S2=12∣−bamp;b2amp;10amp;0amp;1aamp;a2amp;1∣=12ab(a+b)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)

S1S2amp;=16(a+b)312⋅ab(a+b)amp;=13(a+b)2ab=13(ab+ba+2)\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}

amp;∵ba+ab≥2amp;⇒(S1S2)min⁡=43=mnamp;⇒m+n=7\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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