The official answer seems to be wrong. His OE and BE are equal in length. . . . Because the c coordinate (3/2,? 3√3? /2). So OE=3√3? /2，OB=√3 .

So PF=BE=√3? /2? Then everything else is similar, and the final answer is that P is on the image.

According to the nature of central symmetry, AP=BC=? √3, if point P passes through and the X axis of PF ⊥ is at point F, it is easy to know △ APF △ CBE, so AF=CE=3/2, PF=BE=√3? /2? So =3+3/2? =9/2, so P(9/2? ,√3? /2? ) What else? 9/2×√3? /2=? 9√3? /4, so the point p is on the hyperbola.