Area Under Curves Question 2
Question 2 - 2024 (01 Feb Shift 2)
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 $S_{1}$ be the area of the region bounded by the line PQ and the parabola, and $S_{2}$ be the area of the triangle $O P Q$. If the minimum value of $\frac{S_{1}}{S_{2}}$ is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to :
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Answer (7)
Solution
$P Q:-y-a^{2}=\frac{a^{2}-b^{2}}{a+b}(x-a)$
$y-a^{2}=(a-b) x-(a-b) a$
$y=(a-b) x+a b$
$S_{1}=\int_{-b}^{a}\left((a-b) x+a b-x^{2}\right) d x$
$=(a-b) \frac{x^{2}}{2}+(a b) x-\left.\frac{x^{3}}{3}\right|_{-b} ^{a}$
$=\frac{(a-b)^{2}(a+b)}{2}+a b(a+b)-\frac{\left(a^{3}+b^{3}\right)}{3}$
$\frac{S_{1}}{S_{2}}=\frac{\frac{(a-b)^{2}}{2}+a b-\frac{\left(a^{2}+b^{2}-a b\right)}{3}}{\text { athon } \frac{a b}{2}}$
$=\frac{3(a-b)^{2}+6 a b-2\left(a^{2}+b^{2}-a b\right)}{3 a b}$
$=\frac{1}{3}\left[\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}+2\right]$
$=\frac{4}{3}=\frac{\mathrm{m}}{\mathrm{n}} \quad \mathrm{m}+\mathrm{n}=7$
