心似蓝色
幼苗
共回答了18个问题采纳率:83.3% 举报
(1)
![](https://img.yulucn.com/upload/4/cf/4cf66240261608391935e00806f9e69b_thumb.jpg)
,在;(2)
![](https://img.yulucn.com/upload/e/dc/edc38fc47e4a095d5e3a5c76e00c5752_thumb.jpg)
;(3)存在,(
![](https://img.yulucn.com/upload/2/1f/21ff3b2d90558948cf8429c38086ae82_thumb.jpg)
,12).
试题分析:(1)由已知条件先求出C,D两点的坐标,再把其横纵坐标分别代入抛物线的解析式求出b,c,再将点B坐标代入检验即可;(2)BD的长为定值,所以要使△PBD周长最小,只需PB+PD最小,连接DC,则DC与对称轴的交点即为使△PBD周长最小的点;(3)设Q(
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,t)为抛物线对称轴x=
上一点,M在抛物线上,要使四边形BCQM为平行四边形,则BC∥QM且BC=QM,再分①当点M在对称轴的左侧时和①当点M在对称轴的右侧时,讨论即可.
试题解析:(1)∵OA=
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,AD=AC=2
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,∴C(3
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,0),B(
![](https://img.yulucn.com/upload/f/f6/ff618cd04dcf371ba112f41cd2dc1b20_thumb.jpg)
,0).
又在Rt△AOD中,OA=
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,∴OD=
![](https://img.yulucn.com/upload/6/ba/6ba601598ab7a1afbb241dd6f7fefeb8_thumb.jpg)
. ∴D
![](https://img.yulucn.com/upload/8/c3/8c324d851236d0619feba4722c63ae24_thumb.jpg)
.
又∵D,C两点在抛物线上,∴
![](https://img.yulucn.com/upload/5/dc/5dc6466cdc4d928da82f59426bf553ec_thumb.jpg)
,解得
![](https://img.yulucn.com/upload/3/5c/35cabcbbfbbc234d7396cd2c9b7aefff_thumb.jpg)
.
∴抛物线的解析式为
![](https://img.yulucn.com/upload/4/cf/4cf66240261608391935e00806f9e69b_thumb.jpg)
.
又∵当
![](https://img.yulucn.com/upload/3/16/316c50fc6d10439029ae36eaa74f3eae_thumb.jpg)
时,
![](https://img.yulucn.com/upload/9/97/997797c5bc500db77a3ccf74e1ae61ce_thumb.jpg)
,
∴点B(
![](https://img.yulucn.com/upload/f/f6/ff618cd04dcf371ba112f41cd2dc1b20_thumb.jpg)
,0)在该抛物线上.
(2)∵
![](https://img.yulucn.com/upload/f/c2/fc25b3c08e5419a70c68023633e43c1f_thumb.jpg)
,∴抛物线的对称轴方程为:x=
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
.
∵BD的长为定值,∴要使△PBD周长最小,只需PB+PD最小.
连接DC,则DC与对称轴的交点即为使△FBD周长最小的点,
设直线DC的解析式为y=mx+n,
![](https://img.yulucn.com/upload/3/37/337e57be248c79b83bfe8a29a3935b12_thumb.jpg)
,解得
![](https://img.yulucn.com/upload/b/54/b545e79f3ea4e927f11e8c9c37eb4b29_thumb.jpg)
.
∴直线DC的解析式为
![](https://img.yulucn.com/upload/1/40/1404fe1630c79daf24ca9132531bd716_thumb.jpg)
.
在
![](https://img.yulucn.com/upload/1/40/1404fe1630c79daf24ca9132531bd716_thumb.jpg)
中令x=
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
得y=
![](https://img.yulucn.com/upload/d/92/d92119f0d0c6a73f69278d75c13b32ed_thumb.jpg)
. ∴P的坐标为
![](https://img.yulucn.com/upload/e/dc/edc38fc47e4a095d5e3a5c76e00c5752_thumb.jpg)
.
(3)存在,
设Q(
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,t)为抛物线对称轴x=
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
上一点,M在抛物线上,
要使四边形BCQM为平行四边形,则BC∥QM且BC=QM,且点M在对称轴的左侧,
过点Q作直线L∥BC与抛物线交于点M(x,t),由BC=QM得QM=4
![](https://img.yulucn.com/upload/f/69/f6919501a3b5042f142c33565d8354db_thumb.jpg)
,从而x=
![](https://img.yulucn.com/upload/2/1f/21ff3b2d90558948cf8429c38086ae82_thumb.jpg)
,t=12.
故在抛物线上存在点M(
![](https://img.yulucn.com/upload/2/1f/21ff3b2d90558948cf8429c38086ae82_thumb.jpg)
,12)使得四边形BCQM为平行四边形.
1年前
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