题目内容
已知⊙O′过定点A(0,p)(p>0),圆心O′在抛物线C:x2=2py(p>0)上运动,MN为圆O′在x轴上所截得的弦.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240546389032617.jpg)
(1)当O′点运动时,|MN|是否有变化?并证明你的结论;
(2)当|OA|是|OM|与|ON|的等差中项时,试判断抛物线C的准线与圆O′的位置关系,并说明理由.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240546389032617.jpg)
(1)当O′点运动时,|MN|是否有变化?并证明你的结论;
(2)当|OA|是|OM|与|ON|的等差中项时,试判断抛物线C的准线与圆O′的位置关系,并说明理由.
(1)|MN|不变化,其定值为2p 见解析
(2)见解析
(2)见解析
(1)设O′(x0,y0),则x02=2py0(y0≥0),
则⊙O′的半径|O′A|=
,
⊙O′的方程为(x-x0)2+(y-y0)2=x02+(y0-p)2,
令y=0,并把x02=2py0,代入得x2-2x0x+x02-p2=0,
解得x1=x0-p,x2=x0+p,所以|MN|=|x1-x2|=2p,
这说明|MN|不变化,其定值为2p.
(2)不妨设M(x0-p,0),N(x0+p,0).
由题2|OA|=|OM|+|ON|,得2p=|x0-p|+|x0+p|,
所以-p≤x0≤p.
O′到抛物线准线y=-
的距离d=y0+
=
,
⊙O′的半径|O′A|=![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054638918868.png)
=
=![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054639028448.png)
.
因为r>d?x04+4p4>(x02+p2)2?x02<
p2,
又x02≤p2<
p2(p>0),所以r>d,
即⊙O′与抛物线的准线总相交.
则⊙O′的半径|O′A|=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054638918868.png)
⊙O′的方程为(x-x0)2+(y-y0)2=x02+(y0-p)2,
令y=0,并把x02=2py0,代入得x2-2x0x+x02-p2=0,
解得x1=x0-p,x2=x0+p,所以|MN|=|x1-x2|=2p,
这说明|MN|不变化,其定值为2p.
(2)不妨设M(x0-p,0),N(x0+p,0).
由题2|OA|=|OM|+|ON|,得2p=|x0-p|+|x0+p|,
所以-p≤x0≤p.
O′到抛物线准线y=-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054638934422.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054638934422.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054638965709.png)
⊙O′的半径|O′A|=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054638918868.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240546390121149.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054639028448.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054639043687.png)
因为r>d?x04+4p4>(x02+p2)2?x02<
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054639059388.png)
又x02≤p2<
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824054639059388.png)
即⊙O′与抛物线的准线总相交.
![](http://thumb.zyjl.cn/images/loading.gif)
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