SAT1000 - P760

Equations of the asymptotes of the hyperbola are $y=\pm \frac{b}{a}x$ .

Position coordinates of $F_1$ are $(-\sqrt {a^2+b^2},0$ and of $F_2$ are $(\sqrt {a^2+b^2}=0$ .

So, the equation of $\overline {F_2B}$ is $y=b(1+\frac{x}{\sqrt {a^2+b^2}})$ .

Position coordinates of $P$ are $(-\frac{a\sqrt {a^2+b^2}}{a+\sqrt {a^2+b^2}}, \frac{b\sqrt {a^2+b^2}}{a+\sqrt {a^2+b^2}})$ ,

and of $Q$ are $(\frac{a\sqrt {a^2+b^2}}{\sqrt {a^2+b^2}-a}, \frac{b\sqrt {a^2+b^2}}{\sqrt {a^2+b^2}-a})$ .

So, the equation of the perpendicular bisector of $\overline {PQ}$ is

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

Position coordinates of $M$ are $\left (\frac{(a^2+b^2)^{\frac{3}{2}}}{b^2},0\right )$ .

$|\overline {MF_2}|=|\overline {F_1F_2}|\implies 2\sqrt {a^2+b^2}=(\frac{a^2+b^2}{b^2}-1)\sqrt {a^2+b^2}$

$\implies \frac{a^2}{b^2}=2\implies E=\sqrt {1+\frac{b^2}{a^2}}=\sqrt {1.5}\approx 1.2247$ .

Therefore $\lfloor 1000E\rfloor =\boxed {1224}$ .