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共回答了22个问题采纳率:90.9% 举报
(1)∵四边形ABCD是菱形,
∴AB=AD=CD=BC=5,sinB=sinD=
![](https://img.yulucn.com/upload/6/4c/64cc8f0ad6d86e4f218b8aa64fc0062f_thumb.jpg)
;
Rt△OCD中,OC=CD•sinD=4,OD=3;
OA=AD﹣OD=2,即:
A(﹣2,0)、B(﹣5,4)、C(0,4)、D(3,0);
设抛物线的解析式为:y=a(x+2)(x﹣3),得:
2×(﹣3)a=4,a=﹣
![](https://img.yulucn.com/upload/d/05/d05cb66fb392607e465537a0ec302845_thumb.jpg)
;
∴抛物线:y=﹣
![](https://img.yulucn.com/upload/d/05/d05cb66fb392607e465537a0ec302845_thumb.jpg)
x
2 +
![](https://img.yulucn.com/upload/d/05/d05cb66fb392607e465537a0ec302845_thumb.jpg)
x+4.
(2)由A(﹣2,0)、B(﹣5,4)得直线AB:y
1 =﹣
![](https://img.yulucn.com/upload/1/fe/1fe2b59ec6dddf535e5a25a8435da911_thumb.jpg)
x﹣
![](https://img.yulucn.com/upload/0/fa/0faf3ee4ec588b18c525482f1fac8e6d_thumb.jpg)
;
由(1)得:y
2 =﹣
![](https://img.yulucn.com/upload/d/05/d05cb66fb392607e465537a0ec302845_thumb.jpg)
x
2 +
![](https://img.yulucn.com/upload/d/05/d05cb66fb392607e465537a0ec302845_thumb.jpg)
x+4,则:
![](https://img.yulucn.com/upload/9/9c/99c9865fb352ac01af39f3750c28f96f_thumb.jpg)
,解得:
![](https://img.yulucn.com/upload/5/1d/51da9309d091fd9a346a34c6c87fadc9_thumb.jpg)
,
![](https://img.yulucn.com/upload/3/a0/3a0eb0c81d26e6f05958ff2f38f1acb5_thumb.jpg)
;
由图可知:当y
1 <y
2 时,﹣2<x<5.
(3)∵S
△ APE =
![](https://img.yulucn.com/upload/7/61/761b5a024eb2ac998bc3ed5cb36632b2_thumb.jpg)
AE•h,
∴当P到直线AB的距离最远时,S
△ ABC 最大;
若设直线L∥AB,则直线L与抛物线有且只有一个交点时,该交点为点P;
设直线L:y=﹣
![](https://img.yulucn.com/upload/c/20/c2057b9716d1bb45b04cbced93cfe49e_thumb.jpg)
x+b,当直线L与抛物线有且只有一个交点时,
﹣
![](https://img.yulucn.com/upload/b/ab/babe6c3d1e1a616f0396b2f46f6210d3_thumb.jpg)
x+b=﹣
![](https://img.yulucn.com/upload/e/d9/ed93a26cf0a72b93b17d98cc5bdadaf1_thumb.jpg)
x
2 +
![](https://img.yulucn.com/upload/e/d9/ed93a26cf0a72b93b17d98cc5bdadaf1_thumb.jpg)
x+4,且△=0;
求得:b=
![](https://img.yulucn.com/upload/f/f0/ff034619774d7630625bff219523ec2f_thumb.jpg)
,即直线L:y=﹣
![](https://img.yulucn.com/upload/b/ab/babe6c3d1e1a616f0396b2f46f6210d3_thumb.jpg)
x+
![](https://img.yulucn.com/upload/f/f0/ff034619774d7630625bff219523ec2f_thumb.jpg)
;
可得点P(
![](https://img.yulucn.com/upload/3/9d/39d2867c88d85617334dde3306127183_thumb.jpg)
,
![](https://img.yulucn.com/upload/d/2d/d2db939389d7066efbc635ff4f216c58_thumb.jpg)
).
由(2)得:E(5,﹣
![](https://img.yulucn.com/upload/2/f9/2f925b58841a934261d76f59c22a5e69_thumb.jpg)
),则直线PE:y=﹣
![](https://img.yulucn.com/upload/8/34/834651feec0a456d9cc34dd271225f0b_thumb.jpg)
x+9;
则点F(
![](https://img.yulucn.com/upload/2/b5/2b5f1188b1b1da7d17ed74fc2a71ae31_thumb.jpg)
,0),AF=OA+OF=
![](https://img.yulucn.com/upload/e/bf/ebfac861d4854824463d6de36e5ac5b8_thumb.jpg)
;
∴△PAE的最大值:S
△ PAE =S
△ PAF +S
△ AEF =
![](https://img.yulucn.com/upload/7/46/7461a9a45ed2ee2f0b7909f35f1d5ad2_thumb.jpg)
×
![](https://img.yulucn.com/upload/e/bf/ebfac861d4854824463d6de36e5ac5b8_thumb.jpg)
×(
![](https://img.yulucn.com/upload/2/f9/2f925b58841a934261d76f59c22a5e69_thumb.jpg)
+
![](https://img.yulucn.com/upload/2/5b/25b65be3bf2dad9e3060e0225bfe4c47_thumb.jpg)
)=
![](https://img.yulucn.com/upload/7/27/7271b3a7d66246537a4971f9c3f7e8b6_thumb.jpg)
.
综上所述,当P(
![](https://img.yulucn.com/upload/4/23/423cf93be9f1ed0a447ca7dcca62bba0_thumb.jpg)
,
![](https://img.yulucn.com/upload/e/67/e6715ff6c246b5ae038dd6a16a51b305_thumb.jpg)
)时,△PAE的面积最大,为
![](https://img.yulucn.com/upload/2/36/2364e9329135ef3d6aa56ffd1fac4e99_thumb.jpg)
.
(1)由菱形ABCD的边长和一角的正弦值,可求出OC.OD.OA的长,进而确定A.C.D三点坐标,通过待定系数法可求出抛物线的解析式.
(2)首先由A.B的坐标确定直线AB的解析式,然后求出直线AB与抛物线解析式的两个交点,然后通过观察图象找出直线y
1 在抛物线y
2 图象下方的部分.
(3)该题的关键点是确定点P的位置,△APE的面积最大,那么S
△ APE =
![](https://img.yulucn.com/upload/b/c6/bc6f76c29b0a55e7d1eb1a25328f1be2_thumb.jpg)
AE×h中h的值最大,即点P离直线AE的距离最远,那么点P为与直线AB平行且与抛物线有且仅有的唯一交点.
1年前
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