leon866
幼苗
共回答了16个问题采纳率:100% 举报
(1)(0,-3),b=-
![](https://img.yulucn.com/upload/d/63/d63d702263cfb2e72cc3850718e374fa_thumb.jpg)
,c=-3;
(2)由(1),得y=
![](https://img.yulucn.com/upload/5/06/5066f57eae42023a1eda67136b53d65f_thumb.jpg)
x
2 -
![](https://img.yulucn.com/upload/c/2b/c2bc4e3aa9a9c1d5df0573a10743b282_thumb.jpg)
x-3,它与x轴交于A,B两点,得B(4,0),
∴OB=4,又∵OC=3,∴BC=5,
由题意,得△BHP∽△BOC,
∵OC∶OB∶BC=3∶4∶5,
∴HP∶HB∶BP=3∶4∶5,
∵PB=5t,∴HB=4t,HP=3t,
∴OH=OB-HB=4-4t,
由y=
![](https://img.yulucn.com/upload/7/08/7083bc8f7a7a09d3c7a5b751feacb5c6_thumb.jpg)
x-3与x轴交于点Q,得Q(4t,0),
∴OQ=4t,
①当H在Q、B之间时,QH=OH-OQ=(4-4t)-4t=4-8t;
②当H在O、Q之间时,QH=OQ-OH=4t-(4-4t)=8t-4;
综合①,②得QH=|4-8t|;
(3)存在t的值,使以P、H、Q为顶点的三角形与△COQ相似;
①当H在Q、B之间时,QH=4-8t,若
△QHP∽△COQ,则QH∶CO=HP∶OQ,得
![](https://img.yulucn.com/upload/c/a0/ca046be443563e1c57e308fb8bd73d08_thumb.jpg)
,
∴t=
![](https://img.yulucn.com/upload/5/55/555a5e0c9075710d82d0bc9f7dd3750e_thumb.jpg)
,
若△PHQ∽△COQ,则PH∶CO=HQ∶OQ,得
![](https://img.yulucn.com/upload/f/26/f263cc8353a87244f4b34e2e8528025d_thumb.jpg)
,即t
2 +2t-1=0,∴t
1 =
![](https://img.yulucn.com/upload/2/20/220cf8ea82c1783f79aa8aa55c6dd1f2_thumb.jpg)
-1,t
2 =-
![](https://img.yulucn.com/upload/2/20/220cf8ea82c1783f79aa8aa55c6dd1f2_thumb.jpg)
-1(舍去)
②当H在O、Q之间时,QH=8t-4,
若△QHP∽△COQ,则QH∶CO=HP∶OQ,得
![](https://img.yulucn.com/upload/8/7a/87a61a274bb845303ad899e062cb003c_thumb.jpg)
,
∴t=
![](https://img.yulucn.com/upload/8/ac/8acc83d7391c63a86493f4cf0f417897_thumb.jpg)
,
若△PHQ∽△COQ,则PH∶CO=HQ∶OQ,得
![](https://img.yulucn.com/upload/d/53/d539526f871014a2f9befb735b083675_thumb.jpg)
,即t
2 -2t+1=0,∴t
1 =t
2 =1(舍去)
综上所述,存在t的值,t
1 =
![](https://img.yulucn.com/upload/0/b0/0b0ee09d055c2f3c677d839b0d127fef_thumb.jpg)
-1,t
2 =
![](https://img.yulucn.com/upload/d/d8/dd8053c36ca8a6211d6c3c8b6453b5df_thumb.jpg)
,t
3 =
![](https://img.yulucn.com/upload/c/66/c665f6faa23e93af311576cfdd0c031e_thumb.jpg)
。
1年前
5