Three points O(0, 0), P(a, a ^ 2 ), Q(- b, b ^ 2 ), a > 0, b > 0,...

Plotting the diagram of the given data we get,

Area of $△ OPQ$ is given by,

{S}_{2} = \dfrac{1}{2}\left|\begin{cases}0 & 0 & 1 \\ a & {a}^{2} & 1 \\ - b & {b}^{2} & 1\end{cases}\right|

$⇒ {S}_{2} = \dfrac{1}{2}\left(a{b}^{2} + {a}^{2}b\right) . . . \left(i\right)$

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

$⇒ y - {a}^{2} = \left(a - b\right)x - \left(a - b\right)a$

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

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

${S}_{1} = \mathop{∫}_{- b}^{a}\left(\left(a - b\right)x + ab - {x}^{2}\right)dx$

$⇒ {S}_{1} = {\left[\left(a - b\right)\dfrac{{x}^{2}}{2} + \left(ab\right)x - \dfrac{{x}^{3}}{3}\right]}_{- b}^{a}$

$⇒ {S}_{1} = \dfrac{{\left(a - b\right)}^{2}\left(a + b\right)}{2} + ab\left(a + b\right) - \dfrac{\left({a}^{3} + {b}^{3}\right)}{3} . . . \left(iii\right)$

$⇒ \dfrac{{S}_{1}}{{S}_{2}} = \dfrac{\dfrac{{\left(a - b\right)}^{2}}{2} + ab - \dfrac{\left({a}^{2} + {b}^{2} - ab\right)}{3}}{\dfrac{ab}{2}}$

$⇒ \dfrac{{S}_{1}}{{S}_{2}} = \dfrac{3{\left(a - b\right)}^{2} + 6ab - 2\left({a}^{2} + {b}^{2} - ab\right)}{3ab}$

$⇒ \dfrac{{S}_{1}}{{S}_{2}} = \dfrac{3\left({a}^{2} + {b}^{2} - 2ab\right) + 6ab - 2\left({a}^{2} + {b}^{2} - ab\right)}{3ab}$

$⇒ \dfrac{{S}_{1}}{{S}_{2}} = \dfrac{3{a}^{2} + 3{b}^{2} - 6ab + 6ab - 2{a}^{2} - 2{b}^{2} + 2ab}{3ab}$

$⇒ \dfrac{{S}_{1}}{{S}_{2}} = \dfrac{{a}^{2} + {b}^{2} + 2ab}{3ab}$

$⇒ \dfrac{{S}_{1}}{{S}_{2}} = \dfrac{1}{3}$\left[\dfrac{a}{b} + \dfrac{b}{a} + 2\right]$</p><p>$⇒ {\left(\dfrac{{S}_{1}}{{S}_{2}}\right)}_\min = \dfrac{1}{3}\left[1 + 1 + 2\right]$\left\{\text{as}\dfrac{a}{b} + \dfrac{b}{a} \ge 2\right\}$

⇒ {\left(\dfrac{{S}_{1}}{{S}_{2}}\right)}_\min = \dfrac{4}{3}

$⇒ m + n = 7$