As shown in the figure, |OB|=b, |OF 1|=c, so the slope of the straight line PQ =b/c, the slope of the straight line passing through point M and perpendicular to PQ =-c/b, the equation of the straight line PQ is Y=b(x+c)/c, and the equations of the two asymptotes are Y=bx/a respectively. From Y=b(x+c)/c and y=bx/a, the coordinates of point q are [ac/(c-a), BC/(c-a)]; According to y=b(x+c)/c and y=-bx/a, the coordinates of point p are [-ac/(c+a), bc/(c+a)], so the coordinates of the midpoint of the straight line PQ are [square of a * c/(square of c-square of a) and square of b*c/. Therefore, the equation of the straight line passing through point M and perpendicular to PQ is: y-b * square of c/(square of c-a) = (y-b*c/b) * [square of x-a * c/(square of c-a)]. In this equation, let Y=0, and the abscissa of point m is the cube of c/. That is, 3c = the cube of c/(the square of c-the square of a) is simplified as: the square of 2c-the square of 3a =0, that is, the square of c/the square of a =3/2, so c /a = e, so under the root sign, e= (6)/2, so b is selected.