【答案】
分析:解法1:(Ⅰ)依題意,點N的坐標為N(0,-p),可設(shè)A(x
1,y
1),B(x
2,y
2),直線AB的方程為y=kx+p,與x
2=2py聯(lián)立得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/0.png)
消去y得x
2-2pkx-2p
2=0.然后由韋達定理結(jié)合三角形面積公式進行求解.
(Ⅱ)假設(shè)滿足條件的直線l存在,其方程為y=a,AC的中點為O',l與AC為直徑的圓相交于點P,Q,PQ的中點為H,
則O'H⊥PQ,Q'點的坐標為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/1.png)
,y
1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/2.png)
),由此入手能夠求出拋物線的通徑所在的直線.
解法2:(Ⅰ)依題意,點N的坐標為N(0,-p),可設(shè)A(x
1,y
1),B(x
2,y
2),直線AB的方程為y=kx+p,與x
2=2py聯(lián)立得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/3.png)
消去y得x
2-2pkx-2p
2=0.由弦長公式得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/5.png)
,又由點到直線的距離公式得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/6.png)
.由此能求出△ANB面積的最小值.
(Ⅱ)假設(shè)滿足條件的直線l存在,其方程為y=a,則以AC為直徑的圓的方程為(x-0)(x-x
1)-(y-p)(y-y
1)=0,
將直線方程y=a代入得x
2-x
1x+(a-p)(a-y
1)=0,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/7.png)
.由此入手能夠求出拋物線的通徑所在的直線.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/images8.png)
解:法1:(Ⅰ)依題意,點N的坐標為N(0,-p),
可設(shè)A(x
1,y
1),B(x
2,y
2),
直線AB的方程為y=kx+p,與x
2=2py聯(lián)立得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/8.png)
,
消去y得x
2-2pkx-2p
2=0.
由韋達定理得x
1+x
2=2pk,x
1x
2=-2p
2.
于是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/11.png)
,
∴當k=0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/12.png)
.
(Ⅱ)假設(shè)滿足條件的直線l存在,其方程為y=a,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/images14.png)
AC的中點為O',l與AC為直徑的圓相交于點P,Q,PQ的中點為H,
則O'H⊥PQ,Q'點的坐標為(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/13.png)
).
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/15.png)
,
∴|PH|
2=|O'P|
2-|O'H|
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/17.png)
,
∴|PQ|
2=(2|PH|)
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/18.png)
.
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/19.png)
,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/20.png)
,此時|PQ|=p為定值,
故滿足條件的直線l存在,其方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/21.png)
,
即拋物線的通徑所在的直線.
解法2:(Ⅰ)前同解法1,再由弦長公式得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/23.png)
,
又由點到直線的距離公式得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/24.png)
.
從而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/25.png)
,∴當k=0時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/26.png)
.
(Ⅱ)假設(shè)滿足條件的直線l存在,其方程為y=a,則以AC為直徑的圓的方程為(x-0)(x-x
1)+(y-p)(y-y
1)=0,
將直線方程y=a代入得x
2-x
1x+(a-p)(a-y
1)=0,
則|x
1-x
2|
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/27.png)
.
設(shè)直線l與以AC為直徑的圓的交點為P(x
3,y
3),Q(x
4,y
4),
則有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/28.png)
.
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/29.png)
,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/30.png)
,此時|PQ|=p為定值,故滿足條件的直線l存在,其方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181430776538758/SYS201310241814307765387018_DA/31.png)
,
即拋物線的通徑所在的直線.
點評:本小題主要考查直線、圓和拋物線等平面解析幾何的基礎(chǔ)知識,考查綜合運用數(shù)學(xué)知識進行推理運算的能力和解決問題的能力.