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…

Plotting the diagram of the given data we get,

Area of $△OPQ$ is given by,

$S_2=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ a& a^2& 1\\ -b& b^2& 1\end{array}\right|$

$\Rightarrow S_2=\frac{1}{2}\left(a{b}^2+a^{2}b\right) ...\left(i\right)$

Now, equation of line $PQ:y-a^2=\frac{a^2-b^2}{a+b}\left(x-a\right)$

$\Rightarrow y-a^2=\left(a-b\right)x-\left(a-b\right)a$

$\Rightarrow y=\left(a-b\right)x+ab ...\left(ii\right)$

Area bounded by the line $PQ$ and the parabola is given by,

$S_1=\underset{-b}{\overset{a}{\int }}\left(\left(a-b\right)x+ab-x^2\right)dx$

$\Rightarrow S_1={\left[\left(a-b\right)\frac{x^2}{2}+\left(ab\right)x-\frac{x^3}{3}\right]}_{-b}^a$

$\Rightarrow S_1=\frac{{\left(a-b\right)}^2\left(a+b\right)}{2}+ab\left(a+b\right)-\frac{\left(a^3+b^3\right)}{3} ...\left(iii\right)$

$\Rightarrow \frac{S_1}{S_2}=\frac{\frac{{\left(a-b\right)}^2}{2}+ab-\frac{\left(a^2+b^2-ab\right)}{3}}{\frac{ab}{2}}$

$\Rightarrow \frac{S_1}{S_2}=\frac{3{\left(a-b\right)}^2+6ab-2\left(a^2+b^2-ab\right)}{3ab}$

$\Rightarrow \frac{S_1}{S_2}=\frac{3\left(a^2+b^2-2ab\right)+6ab-2\left(a^2+b^2-ab\right)}{3ab}$

$\Rightarrow \frac{S_1}{S_2}=\frac{3a^2+3b^2-6ab+6ab-2a^2-2b^2+2ab}{3ab}$

$\Rightarrow \frac{S_1}{S_2}=\frac{a^2+b^2+2ab}{3ab}$

$\Rightarrow \frac{S_1}{S_2}=\frac{1}{3}\left[\frac{a}{b}+\frac{b}{a}+2\right]$

$\Rightarrow {\left(\frac{S_1}{S_2}\right)}_{\min }=\frac{1}{3}\left[1+1+2\right]$ $\left\{\text{as} \frac{a}{b}+\frac{b}{a}\ge 2\right\}$

$\Rightarrow {\left(\frac{S_1}{S_2}\right)}_{\min }=\frac{4}{3}$

$\Rightarrow m+n=7$