foxzmb
春芽
共回答了16个问题采纳率:75% 举报
(1)证明见解析;(2)y=-
![](https://img.yulucn.com/upload/8/0b/80bb8af936ad65ed3d067b5573a661b0_thumb.jpg)
x
2 -
![](https://img.yulucn.com/upload/d/5d/d5d8d3291ba700c5b7ace3fb3a527506_thumb.jpg)
x+4;顶点E是否在直线CD上,理由见解析;P
1 (-10,-6),P
2 (10,-36).
试题分析:(1)连接O′C,由CD是⊙O的切线,可得O′C⊥CD,则可证得O′C∥AD,又由O′A=O′C,则可证得∠CAD=∠CAB;
(2)①首先证得△CAO∽△BCO,根据相似三角形的对应边成比例,可得OC
2 =OA•OB,又由tan∠CAO=tan∠CAD=
![](https://img.yulucn.com/upload/c/f9/cf968ecd0ff164c851d53325ba663b49_thumb.jpg)
,则可求得CO,AO,BO的长,然后利用待定系数法即可求得二次函数的解析式;
②首先证得△FO′C∽△FAD,由相似三角形的对应边成比例,即可得到F的坐标,求得直线DC的解析式,然后将抛物线的顶点坐标代入检验即可求得答案;
③根据题意分别从PA∥BC与PB∥AC去分析求解即可求得答案,小心漏解.
试题解析:(1)证明:连接O′C,
∵CD是⊙O′的切线,
∴O′C⊥CD,
∵AD⊥CD,
∴O′C∥AD,
∴∠O′CA=∠CAD,
∵O′A=O′C,
∴∠CAB=∠O′CA,
∴∠CAD=∠CAB;
(2)①∵AB是⊙O′的直径,
∴∠ACB=90°,
∵OC⊥AB,
∴∠CAB=∠OCB,
∴△CAO∽△BCO,
∴
![](https://img.yulucn.com/upload/4/4e/44eb9ba14134f22e769596d75cc9fe30_thumb.jpg)
,
即OC
2 =OA•OB,
∵tan∠CAO=tan∠CAD=
![](https://img.yulucn.com/upload/c/f9/cf968ecd0ff164c851d53325ba663b49_thumb.jpg)
,
∴AO=2CO,
又∵AB=10,
∴OC
2 =2CO(10-2CO),
解得CO
1 =4,CO
2 =0(舍去),
∴CO=4,AO=8,BO=2
∵CO>0,
∴CO=4,AO=8,BO=2,
∴A(-8,0),B(2,0),C(0,4),
∵抛物线y=ax
2 +bx+c过点A,B,C三点,
∴c=4,
由题意得:
![](https://img.yulucn.com/upload/b/59/b59bc5e60ce0601046e97bddaab580f6_thumb.jpg)
,
解得:
![](https://img.yulucn.com/upload/b/eb/beb962e892e332c47e617e2b42638bcb_thumb.jpg)
,
∴抛物线的解析式为:y=-
![](https://img.yulucn.com/upload/8/0b/80bb8af936ad65ed3d067b5573a661b0_thumb.jpg)
x
2 -
![](https://img.yulucn.com/upload/d/5d/d5d8d3291ba700c5b7ace3fb3a527506_thumb.jpg)
x+4;
②设直线DC交x轴于点F,
∴△AOC≌△ADC,
∴AD=AO=8,
∵O′C∥AD,
∴△FO′C∽△FAD,
∴
![](https://img.yulucn.com/upload/2/27/2276a1cf38498a4c74e32051f0111a6f_thumb.jpg)
,
∴O′F•AD=O′C•AF,
∴8(BF+5)=5(BF+10),
∴BF=
![](https://img.yulucn.com/upload/2/a6/2a68cc1c8a3599b72421f60bdfd04cb8_thumb.jpg)
,F(
![](https://img.yulucn.com/upload/e/b6/eb6c939f6098fd5e1eba1590634c42e6_thumb.jpg)
,0);
设直线DC的解析式为y=kx+m,
则
![](https://img.yulucn.com/upload/4/f2/4f2cca4788b862cbe209405042eeeb28_thumb.jpg)
,
解得:
![](https://img.yulucn.com/upload/2/9e/29e879b35355ad57666c7ace552acf6f_thumb.jpg)
,
∴直线DC的解析式为y=-
![](https://img.yulucn.com/upload/2/9d/29d3f515ac014d4989c28f9e149cc939_thumb.jpg)
x+4,
由y=-
![](https://img.yulucn.com/upload/8/0b/80bb8af936ad65ed3d067b5573a661b0_thumb.jpg)
x
2 -
![](https://img.yulucn.com/upload/d/5d/d5d8d3291ba700c5b7ace3fb3a527506_thumb.jpg)
x+4=-
![](https://img.yulucn.com/upload/8/0b/80bb8af936ad65ed3d067b5573a661b0_thumb.jpg)
(x+3)
2 +
![](https://img.yulucn.com/upload/0/28/028928ddf82c5878892d3c0bf71d3f20_thumb.jpg)
得顶点E的坐标为(-3,
![](https://img.yulucn.com/upload/0/28/028928ddf82c5878892d3c0bf71d3f20_thumb.jpg)
),
将E(-3,
![](https://img.yulucn.com/upload/0/28/028928ddf82c5878892d3c0bf71d3f20_thumb.jpg)
)代入直线DC的解析式y=--
![](https://img.yulucn.com/upload/2/9d/29d3f515ac014d4989c28f9e149cc939_thumb.jpg)
x+4中,
右边=-
![](https://img.yulucn.com/upload/2/9d/29d3f515ac014d4989c28f9e149cc939_thumb.jpg)
×(-3)+4=
![](https://img.yulucn.com/upload/0/28/028928ddf82c5878892d3c0bf71d3f20_thumb.jpg)
=左边,
∴抛物线顶点E在直线CD上;
(3)存在,P
1 (-10,-6),P
2 (10,-36).
①∵A(-8,0),C(0,4),
∴过A、C两点的直线解析式为y=
![](https://img.yulucn.com/upload/c/f9/cf968ecd0ff164c851d53325ba663b49_thumb.jpg)
x+4,
设过点B且与直线AC平行的直线解析式为:y=
![](https://img.yulucn.com/upload/c/f9/cf968ecd0ff164c851d53325ba663b49_thumb.jpg)
x+b,把B(2,0)代入得b=-1,
∴直线PB的解析式为y=
![](https://img.yulucn.com/upload/c/f9/cf968ecd0ff164c851d53325ba663b49_thumb.jpg)
x-1,
∴
![](https://img.yulucn.com/upload/c/28/c284c43d1d4533f47afca6358f3f6d3f_thumb.jpg)
,
解得
![](https://img.yulucn.com/upload/6/27/62745bc168f5dcb5b93a5eeac31eb63d_thumb.jpg)
,
![](https://img.yulucn.com/upload/f/7d/f7d244e7b2c9435f2584492e7fbe3fb8_thumb.jpg)
(舍去),
∴P
1 (-10,-6).
②求P
2 的方法应为过点A作与BC平行的直线,
可求出BC解析式,进而求出与之平行的直线的解析式,
与求P
1 同法,可求出x
1 =-8,y
1 =0(舍去);x
2 =10,y
2 =-36.
∴P
2 的坐
1年前
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