Parth's Parabola

Let the point of tangency of the line with $S$ be $P(a,b).$ Then since $\frac{dy}{dx} = -2x$ the equation of the line will be

$y - b = -2a(x - a) \Longrightarrow y - (2 - a^{2}) = -2ax + 2a^{2} \Longrightarrow y = a^{2} - 2ax + 2.$

Then for the $x$ -intercept we have $0 = a^{2} - 2ax + 2 \Longrightarrow x = \dfrac{a^{2} + 2}{2a}.$

For the $y$ -intercept we have $y = a^{2} + 2.$

The length $D$ of the tangent line segment is then

$D = \sqrt{\dfrac{(a^{2} + 2}{4a^{2}} + (a^{2} + 2)^{2}} = \dfrac{a^{2} + 2}{2a}\sqrt{1 + 4a^{2}}.$

Differentiating $D$ with respect to $a$ and setting the result equal to $0$ gives us

$\dfrac{dD}{da} = \dfrac{8a^{4} + a^{2} - 2}{2a^{2}\sqrt{1 + 4a^{2}}} = 0 \Longrightarrow 8a^{4} + a^{2} - 2 = 0 \Longrightarrow a^{2} = \dfrac{-1 + \sqrt{65}}{16},$

where we took the positive root since we must have $a^{2} \ge 0.$

Thus $a = \dfrac{\sqrt{\sqrt{65} - 1}}{4},$ at which value we have

$D_{min} = L = \dfrac{\sqrt{65} + 31}{16\sqrt{\sqrt{65} - 1}}\sqrt{\sqrt{65} + 3} = \dfrac{\sqrt{1026 + 62\sqrt{65}}\sqrt{17 + \sqrt{65}}}{64} =$

$\dfrac{21472 + 2080\sqrt{65}}{64} = \sqrt{\dfrac{671 + 65\sqrt{65}}{128}}.$

Thus $m + n = 671 + 65 = \boxed{736}.$

(Note that in simplifying the radical expression for $L$ I multiplied top and bottom by $\sqrt{\sqrt{65} + 1}$ and rewrote $\sqrt{65} + 31$ as $\sqrt{1026 + 62\sqrt{65}}.$

Also, since the tangent line segment can be made arbitrarily long by choosing the point $P$ arbitrarily close to the $y$ -axis, we can rest assured that the value found via the above method is a minimum.)