Rectangle on a Parabola

Let the coordinates of $P$ be $(p, p^2)$ and $R$ be $(r, r^2)$ . Without loss of generality, let $p > 0$ , $r < 0$ , and $PS = 2RS$ .

Then the tangent line through $P$ is $y = 2px - p^2$ and the tangent line through $R$ is $y = 2rx - r^2$ .

As sides of a rectangle, the two tangent lines must be perpendicular, so $2p \cdot 2r = -1$ , or $r = -\frac{1}{4p}$ . Therefore, the coordinates of $R$ are $(-\frac{1}{4p}, \frac{1}{16p^2})$ and the tangent line through $R$ is $y = -\frac{1}{2p}x - \frac{1}{16p^2}$ .

The coordinates of the intersection of the two tangent lines at $S$ is then $(\frac{4p^2 - 1}{8p}, -\frac{1}{4})$ .

By the distance equation, $PS = \frac{\sqrt{(4p^2 + 1)^3}}{8p}$ and $RS = \frac{\sqrt{(4p^2 + 1)^3}}{16p^2}$ .

Since $PS = 2RS$ , $\frac{\sqrt{(4p^2 + 1)^3}}{8p} = 2 \cdot \frac{\sqrt{(4p^2 + 1)^3}}{16p^2}$ , which solves to $p = 1$ for $p > 0$ .

The area of the rectangle is $A_{PQRS} = PS \cdot RS = \frac{\sqrt{(4p^2 + 1)^3}}{8p} \cdot \frac{\sqrt{(4p^2 + 1)^3}}{16p^2} = \frac{(4p^2 + 1)^3}{128p^3}$ . Since $p = 1$ , $A_{PQRS} = \frac{(4 \cdot 1^2 + 1)^3}{128 \cdot 1^3} = \frac{125}{128}$ .

Therefore, $a = 125$ , $b = 128$ , and $a + b = \boxed{253}$ .

Let $P(t,t^2),R(u,u^2)$ be the given points on the parabola. Then position coordinates of $S$ are $(\frac{t+u}{2},tu)$ .

Tangents at $P$ and $R$ are mutually perpendicular $\implies tu=-\dfrac {1}{4}$

$|\overline {PS}|=2|\overline {RS}|\implies t^2-4u^2=\dfrac {3}{4}$

$\implies t^2-\dfrac {1}{4t^2}=\dfrac {3}{4}\implies 4t^4-3t^2-1=0$

$\implies t=1,u=-\dfrac {1}{4},|\overline {PS}|=\dfrac {5\sqrt 5}{8}$ .

So, the area of the rectangle is $\dfrac {1}{2}\times |\overline {PS}|^2=\dfrac {125}{128}$

Hence $a=125,b=128,a+b=\boxed {253}$ .