如图,直角梯形OABC中,OC∥AB,C(0,3),B(4,1),以BC为直径的圆交x轴于E,D两点(D点在E点右方)。(1)求点E,D的坐标;(2)求过B,C,D三点的抛物线的函数关系式;(3)过B,C,-九年级数学

题文

如图,直角梯形OABC中,OC∥AB,C(0,3),B(4,1),以BC为直径的圆交x轴于E,D两点(D点在E点右方)。

(1)求点E,D 的坐标;
(2)求过B,C,D三点的抛物线的函数关系式;
(3)过B,C,D三点的抛物线上是否存在点Q,使△BDQ是以BD为直角边的直角三角形?若不存在,说明理由;若存在,求出点Q的坐标。
题型:解答题  难度:偏难

答案

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解:(1)在BC上取中点G,并过G作GH⊥x轴于H,连接GD


∴H(2,0)
,GH=2-0=2
又DG=BG=

∴D(3,0),E(1,0)。
(2)设过B、C、D三点的抛物线表达式为

解得

(3)设Q,由(2)可得Q
过Q作QN⊥x轴于N
分2种情况:
①当∠BDQ=90°时,
∴∠NDQ+∠BDA=90°
∵∠DNQ=∠BAD=90°
∴∠NDQ+∠NQD=90°
∴∠NQD=∠BDA
∴△NDQ∽△ABD


解得
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