【答案】
分析:(1)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/0.png)
在拋物線上,得p=2,由此能導(dǎo)出拋物線的焦點F的坐標(biāo)和準(zhǔn)線l的方程.
(2)拋物線的方程為y
2=4x,過焦點F(1,0)且傾斜角為60°的直線m的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/1.png)
,由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/2.png)
可得3x
2-10x+3=0,解得點A、B的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/4.png)
,由此能導(dǎo)出k
MA、k
MF、k
MB成等差數(shù)列.
(3)①推廣命題:若拋物線的方程為y
2=4x,過焦點F的直線m交拋物線于A、B兩點,M為拋物線準(zhǔn)線上的一點,直線MA、MF、MB的斜率分別記為k
MA、k
MF、k
MB,則k
MA、k
MF、k
MB成等差數(shù)列.再由拋物線的性質(zhì)和韋達定理進行證明.
②推廣命題:若拋物線的方程為y
2=2px(p>0),過焦點F的直線m交拋物線于A、B兩點,M為拋物線準(zhǔn)線上的一點,直線MA、MF、MB的斜率分別記為k
MA、k
MF、k
MB,則k
MA、k
MF、k
MB成等差數(shù)列.再由拋物線的性質(zhì)結(jié)合分類討論思想進行證明.
解答:解:(1)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/5.png)
在拋物線上,由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/6.png)
得p=2
∴拋物線的焦點坐標(biāo)為F(1,0),
準(zhǔn)線l的方程為x=-1
(2)證明:∵拋物線的方程為y
2=4x,過焦點F(1,0)且傾斜角為60°的直線m的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/7.png)
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/8.png)
可得3x
2-10x+3=0
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/9.png)
解得點A、B的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/11.png)
∵拋物線的準(zhǔn)線方程為x=-1,設(shè)點M的坐標(biāo)為M(-1,t),
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/13.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/14.png)
,
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/15.png)
知k
MA、k
MF、k
MB成等差數(shù)列.
(3)本小題可根考生不同的答題情況給予評分
①推廣命題:若拋物線的方程為y
2=4x,過焦點F的直線m交拋物線于A、B兩點,M為拋物線準(zhǔn)線上的一點,直線MA、MF、MB的斜率分別記為k
MA、k
MF、k
MB,則k
MA、k
MF、k
MB成等差數(shù)列.
證明:
拋物線y
2=4x的焦點坐標(biāo)為F(1,0),當(dāng)直線l
1平行于y軸時,
由(2)知命題成立.
設(shè)M點坐標(biāo)為M(-1,t)
當(dāng)直線m不平行于y軸時,設(shè)m的方程為y=k(x-1),其與拋物線的交點坐標(biāo)為A(x
1,y
1)、B(x
2,y
2),則有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/17.png)
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/18.png)
得ky
2-4y-4k=0,即y
1y
2=-4
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/20.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/23.png)
,∴k
MA+k
MB=2k
MF,即k
MA、k
MF、k
MB成等差數(shù)列
②推廣命題:若拋物線的方程為y
2=2px(p>0),過焦點F的直線m交拋物線于A、B兩點,M為拋物線準(zhǔn)線上的一點,直線MA、MF、MB的斜率分別記為k
MA、k
MF、k
MB,則k
MA、k
MF、k
MB成等差數(shù)列.
證明:拋物線的焦點F的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/24.png)
,準(zhǔn)線方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/25.png)
,設(shè)M點坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/26.png)
設(shè)m與拋物線的交點坐標(biāo)為A(x
1,y
1)、B(x
2,y
2),則有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/27.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/28.png)
(ⅰ)當(dāng)直線m平行于y軸時,直線m的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/29.png)
,
此時有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/30.png)
,∴y
1y
2=-p
2(ⅱ)當(dāng)直線m不平行于y軸時,直線m的方程可設(shè)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/31.png)
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/32.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/33.png)
∴y
1y
2=-p
2![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/34.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/35.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/37.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024183340234132153/SYS201310241833402341321022_DA/38.png)
,
∴k
MA+k
MB=2k
MF,即k
MA、k
MF、k
MB成等差數(shù)列
點評:本題主要考查直線與圓錐曲線的綜合應(yīng)用能力,具體涉及到軌跡方程的求法及直線與拋物線的相關(guān)知識,解題時要注意合理地進行等價轉(zhuǎn)化.