小麻雀4802
幼苗
共回答了13个问题采纳率:84.6% 举报
(1)由y=
![](https://img.yulucn.com/upload/1/d5/1d5e0f20f0818f25c229a25e02a83bed_thumb.jpg)
(x
2 ―8x―180),令y=0,
得x
2 ―8x―180=0,(x―18)(x+10)=0,∴x=18,x=―10,
∴A(18,0)……………………………………………………………………………………1分
在y=
![](https://img.yulucn.com/upload/1/d5/1d5e0f20f0818f25c229a25e02a83bed_thumb.jpg)
x
2 ―
![](https://img.yulucn.com/upload/3/ce/3ceb594b799b254d28378521ac0ecdb9_thumb.jpg)
x―10,令x=0,y=―10,即B(0,―10).……………………………2分
∵BC∥OA,故点C的纵坐标为―10.
由―10 y=
![](https://img.yulucn.com/upload/1/d5/1d5e0f20f0818f25c229a25e02a83bed_thumb.jpg)
x
2 ―
![](https://img.yulucn.com/upload/3/ce/3ceb594b799b254d28378521ac0ecdb9_thumb.jpg)
x―10,得x=8或x=0 .
即C(8,―10)且易求出顶点
![](https://img.yulucn.com/upload/3/c5/3c59c0811de25f2f75817db95328c5e4_thumb.jpg)
……………………………………………………3分
于是A(18,0),B(0,―10),C(8,―10).顶点坐标为
![](https://img.yulucn.com/upload/f/87/f87511ab63392c3186d4f08b4f244631_thumb.jpg)
……………4分
(2)若四边形PQCA为平行四边形,
![](https://img.yulucn.com/upload/8/d6/8d63f50e4f957daea66f050a73a166bf_thumb.jpg)
由于QC∥PA,故只要QC=PA即可,
而PA=18―QC=t,故18―4t=t,得
![](https://img.yulucn.com/upload/0/ee/0ee4535a974404307136823ac60045a3_thumb.jpg)
………………………………………6分
(3)设点P运动t秒,则OP=4t,QC=t,0<t<4.5.说明P在线段OA上,且不与
点O,A重合.
∵QC∥OP,∴△QDC∽△PDO,∴
同理QC∥AF,故
![](https://img.yulucn.com/upload/2/75/2752e646abfd99b9dad37106bc82eec9_thumb.jpg)
即
∴AF=4t=OP.∴PF=PA+AF=PA+OP=18,………………………………………7分
∵点Q到直线PF的距离d=10,
∴S
△ PQF =
![](https://img.yulucn.com/upload/d/81/d81bbf0d0200aabf07613218e570a414_thumb.jpg)
••PF•d=
![](https://img.yulucn.com/upload/d/81/d81bbf0d0200aabf07613218e570a414_thumb.jpg)
×18×10=90.
所以△PQF的面积总为定值90.…………………………………………………………9分
(4)故当
![](https://img.yulucn.com/upload/5/61/56154e7aae53641ead87746f8148d820_thumb.jpg)
不存在等腰三角形△PQF.……………………………………10分
当
![](https://img.yulucn.com/upload/8/35/835570fe7f3274f9f093c30da1024bfd_thumb.jpg)
时,
![](https://img.yulucn.com/upload/e/35/e35cf1437adf27e62c97ff5db571a64a_thumb.jpg)
为等腰三角形.…………………………………………12分
略
1年前
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