【答案】
分析:(1)根據(jù)已知,B點的縱坐標等于A點的縱坐標.OC=OB可知A、B在以坐標原點O為圓心,以O(shè)C長為半徑的圓上,即可得出B點坐標;
(2)①首先根據(jù)已知條件,求出OB的一次函數(shù)解析式.進而確定出P點、H點的坐標.分別用t表示△OPH與△OBC的面積,再根據(jù)已知條件△OPH的面積等于△OBC面積的
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/0.png)
,列出關(guān)于t的一元一次方程式.解方程即可求出t的值.
②首先確定B點、P點的坐標.再根據(jù)△OBP分別以O(shè)P、OB為底邊的面積求法,列出關(guān)于t的等量關(guān)系式,解t即可.
解答:解:(1)∵AB∥OC,OC=OB,A的坐標為(0,8),點C的坐標為(10,0)
∴對B點,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/1.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/2.png)
;
∴B(6,8);
(2)①分為兩種情況:如圖1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/images3.png)
CP=t,OP=10-t,
設(shè)直線CB的解析式是y=kx+b,
C(10,0)B(8,6)代入得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/3.png)
,
解得:k=-3,b=30,
∴y=-3x+30,
把x=10-t代入得:y=-3(10-t)+30=3t,
∴PH=3t,
即S
△OPH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/4.png)
OP•PH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/5.png)
×(10-t)×3t,
假如存在某個時刻,使△OPH的面積等于△OBC面積的
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/6.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/7.png)
×(10-t)×3t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/8.png)
×8×10
3t
2-30t+80=0,
b
2-4ac=900-960<0,
此方程無解,
即此時不存在某個時刻,使△OPH的面積等于△OBC面積的
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/9.png)
;
如圖2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/images11.png)
由(1)可知,正比例函數(shù)OB的解析式是y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/10.png)
,
OP=10-t,
把x=10-t代入y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/11.png)
x得:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/12.png)
(10-t)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/13.png)
,
即PH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/14.png)
,
S
△OPH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/15.png)
OP•PH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/16.png)
×(10-t)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/17.png)
(10-t)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/18.png)
(10-t)
2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/20.png)
=40,
假設(shè)存在某個時刻,使△OPH的面積等于△OBC面積的
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/21.png)
,
即:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/22.png)
(10-t)
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/23.png)
×40.
解得t=7,經(jīng)檢驗符合題意,
所以存在某個時刻,使△OPH的面積等于△OBC面積的
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/24.png)
,此時t=7;
②第一種情況:如圖2,連接PB,OB與圓P相切,切點為K,PC=t,
由(1)知B點的坐標為(6,8),
OB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/25.png)
=10,
P點的坐標為(10-t,0),
對△OBP,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/26.png)
,即(10-t)×8=10×PK,
∴PK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/27.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/images30.png)
又∵PK、PC均為⊙P的半徑,
∴PK=PC,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/28.png)
=t,
解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/29.png)
.
所以,當⊙P與線段OB只有一個公共點時,t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/30.png)
.
第二種情況:當t=10時,P和O重合,此時PB=PC,即B在⊙P上、O在⊙P內(nèi),且⊙P和OB只有一個交點B;
當t=5時,O在⊙P上,當t>5時,O在⊙P內(nèi),
∴當5<t≤10時,⊙P和OB只有一個交點;
即當t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202654296365051/SYS201311032026542963650026_DA/31.png)
或5<t≤10時,⊙P和OB只有一個交點.
點評:本題考查了直線與圓的位置關(guān)系、三角形面積的計算.解決①的關(guān)鍵是根據(jù)已知條件及三角形的面積列出關(guān)于t的一元一次方程式,進而得出結(jié)果;解決②的關(guān)鍵是將計算長度轉(zhuǎn)化為同一三角形從不同角度來看面積的計算,進而求出未知長度.