∴Rt△POE∽Rt△BPA
That is, y = (0 and when x=2, y has the maximum value.
(2) It is known that both △ PAB and △ Poe are isosceles triangles, so we can get p (1, 0), e (0, 1), b (4 4,3) ...
Let the parabola passing through these three points be y=ax2+bx+c, then ⅷ
y
(3) From (2), it can be known that ∠ EPB = 90, that is, the conditions are met when point Q and point B coincide.
The straight line PB is y=x- 1, and intersects the y axis at the point (0,-1).
Translate PB upward by 2 units through point E(0, 1).
The straight line is y=x+ 1.
From ∴Q(5,6).
So there are two points Q (4 4,3) and (5,6) on the parabola that satisfy the condition.