Because PF coincides with PD, PE is the center line of angle OPD, because angle EPD is equal to 180 degrees minus angle DPA, and PB is the center line of angle DPA, so angle BPA is equal to half of angle DPA, and because triangle BPA is a right triangle, angle PBA is equal to 90 degrees minus angle BPA, which is equal to angle EPO, because OE = Y, OP = X and AP = 4-X.

Because the opening of this image is downward and the vertex is (0,4/3), the maximum value of y is 4/3.

If point D is on the side of BC, BD=BA, that is to say, quadrilateral PABD is a square with a side length of 3, that is, PD is perpendicular to OA, so the coordinate of point P is (1, 0). Since OP=PF, quadrilateral EOPF is a square with a side length of 1, the coordinate of point E is (0,/kloc.

(I heard that 2 is a square symbol) So let the equation be y = AX 2+BX+C, and substitute three points respectively to get the values of A, B and C, and then return to the equation.

There is no such point q, because if PE is a right-angled side, that is, the angle EQP is a right angle. Such a point exists only on a circle with a diameter of EP and a center of EP. Only in this case will there be a right angle, and this circle has two intersections with parabola, that is, E and P. A circle can have at most four points with parabola, but the shape of this parabola has been constrained by point B, and the abscissa of its axis of symmetry is greater than point P. That is to say, both EP points are on the left side of the parabola, and the diameter of the circle is limited, but point P is not the vertex, so such a circle and parabola can have at most two points, and there will be no third one.

I'm sorry for the confusion, but there's probably no problem with the thinking!