उत्तर (7)

समाधान

$P Q:y-a^{2}=\dfrac{a^{2}-b^{2}}{a+b}(x-a)$

$y-a^{2}=(a-b) x-(a-b) a$

$y=(a-b) x+a b$

$S _1=\int _{-b}^{a}\left((a-b) x+a b-x^{2}\right) d x$

$=(a-b) \dfrac{x^{2}}{2}+(a b) x-\left.\dfrac{x^{3}}{3}\right| _{-b} ^{a}$

$=\dfrac{(a-b)^{2}(a+b)}{2}+a b(a+b)-\dfrac{\left(a^{3}+b^{3}\right)}{3}$

$\dfrac{S _1}{S _2}=\dfrac{\dfrac{(a-b)^{2}}{2}+a b-\dfrac{\left(a^{2}+b^{2}-a b\right)}{3}}{ \dfrac{a b}{2}}$

$=\dfrac{3(a-b)^{2}+6 a b-2\left(a^{2}+b^{2}-a b\right)}{3 a b}$

$=\dfrac{1}{3}\left[\dfrac{a}{b}+\dfrac{b}{a}+2\right]$

AM $\geq$ GM का उपयोग करते हुए ($\because a,b>0$)

$\begin{aligned} & \frac{\frac{a}{b}+\frac{b}{a}}{2} \geqslant \sqrt{\frac{a}{b} \times \frac{b}{a}} \ & \Rightarrow \frac{a}{b}+\frac{b}{a} \geqslant 2 .\end{aligned}$

$ \Rightarrow \dfrac{1}{3}[2+2]$

$=\dfrac{4}{3}=\dfrac{m}{n}$

$\Rightarrow m+n=7$