Mathematics examination questions and answers for senior high school entrance examination

Solution: (1) If ∠ APD and ∠OPF are equally divided by known PB, then ∠ BPE = 90. ∴ OPE+∠ APB = 90。 And ∞

∴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

(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.