Dynamic Geometry: P13

Label the diagram as follows, and let the coordinates of $P$ be $(p, p^2)$ and $R$ be $(r, r^2)$ .

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 $r = -\cfrac{1}{4p}$ . Therefore, the coordinates of $R$ are $(-\cfrac{1}{4p}, \cfrac{1}{16p^2})$ and the tangent line through $R$ is $y = -\cfrac{1}{2p}x - \cfrac{1}{16p^2}$ .

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

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

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

The center $M$ of the rectangle is the midpoint between $P$ and $R$ , so its coordinates are $(\cfrac{4p^2 - 1}{8p}, \cfrac{16p^4 + 1}{32p^2})$ .

When the area of the rectangle is $\cfrac{(4p^2 + 1)^3}{128p^3} = \cfrac{125}{128}$ , $p = 1$ , which means $M$ is at $(\cfrac{3}{8}, \cfrac{17}{32})$ on the pink segment, and when the area of the rectangle is $\cfrac{(4p^2 + 1)^3}{128p^3} = \cfrac{125}{54}$ , $p = \cfrac{3}{2}$ , which means $M$ is at $(\cfrac{2}{3}, \cfrac{41}{36})$ on the orange segment.

Since $x = \cfrac{4p^2 - 1}{8p}$ can be rearranged to $p = x + \cfrac{1}{2}\sqrt{4x^2 + 1}$ , the locus of points for $M$ is $y = \cfrac{16(x + \frac{1}{2}\sqrt{4x^2 + 1})^4 + 1}{32(x + \frac{1}{2}\sqrt{4x^2 + 1})^2} = 2x^2 + \cfrac{1}{4}$ .

Therefore, the bounded area is $A = \int_{\frac{3}{8}}^{\frac{2}{3}} (2x^2 + \cfrac{1}{4} - x^2) dx = \cfrac{6391}{41472}$ , so that $a = 6391$ , $b = 41472$ , and $b - a = \boxed{35081}$ .