在平面直角坐標系xOy中,點P到點F(3,0)的距離的4倍與它到直線x=2的距離的3倍之和記為d,當P點運動時,d恒等于點P的橫坐標與18之和
(Ⅰ)求點P的軌跡C;
(Ⅱ)設過點F的直線I與軌跡C相交于M,N兩點,求線段MN長度的最大值.
【答案】
分析:(1)由題意,要求動點的軌跡方程,由于已經(jīng)告訴了動點所滿足的約束條件所以利用直接法求其軌跡即可:
(2)由題意及解析式畫出圖形,利用直線與曲線的軌跡方程聯(lián)立,通過圖形討論直線與軌跡的交點,利用兩點間的距離公式求解即可.
解答:解(Ⅰ)設點P的坐標為(x,y),由題設則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/0.png)
3︳x-2︳①由題意軌跡圖(1)如下:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/images1.png)
(圖1)
當x>2時,由①得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/1.png)
,
化簡得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/2.png)
.
當x≤2時由①得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/3.png)
化簡得y
2=12x
故點P的軌跡C是橢圓
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/4.png)
在直線x=2的右側
部分與拋物線C
2:y
2=12x在直線
x=2的左側部分(包括它與直線x=2
的交點)所組成的曲線,參見圖1
(Ⅱ)如圖2所示,
易知直線x=2與C
1,C
2的交點都是A(2,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/5.png)
),
B(2,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/6.png)
),直線AF,BF的斜
率分別為k
AF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/7.png)
,k
BF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/8.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/images10.png)
圖2
當點P在C
1上時,由②知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/9.png)
.④
當點P在C
2上時,由③知|PF|=3+x⑤
若直線l的斜率k存在,則直線l的方程為y=k(x-3)
(1)當k≤k
AF,或k≥k
BF,即k≤-2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/10.png)
時,直線I與軌跡C的兩個交點M(x
1,y
1),N(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/12.png)
)都在C
1上,此時由④知
|MF|=6-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/13.png)
x
1|NF|=6-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/15.png)
從而|MN|=|MF|+|NF|=(6-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/16.png)
x
1)+(6-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/18.png)
)=12-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/19.png)
(x
1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/20.png)
)
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/21.png)
得(3+4k
2)x
2-24k
2x+36k
2-108=0則x
1,x是這個方程的兩根,
所以x
1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/23.png)
*|MN|=12-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/24.png)
(x
1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/25.png)
)=12-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/26.png)
因為當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/27.png)
,或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/28.png)
時,k
2≥24,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/29.png)
.
當且僅當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/30.png)
時,等號成立.
(2)當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/31.png)
時,直線L與軌跡C的兩個交點M(x
1,y
1),N(x
2,y
2)分別在C
1,C
2上,不妨設點M在C
1上,點C
2上,則④⑤知,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/32.png)
設直線AF與橢圓C
1的另一交點為E(x
,y
),則x
<x
1,x
2<2.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/33.png)
所以|MN|=|MF|+|NF|<|EF|+|AF|=|AE|.而點A,E都在C
1上,且
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/34.png)
,有(1)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/35.png)
若直線ι的斜率不存在,則x
1=x
2=3,此時
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/36.png)
綜上所述,線段MN長度的最大值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214527137856395/SYS201310232145271378563019_DA/37.png)
.
點評:(1)此問重點考查了直接法求動點的軌跡方程,還考查了對于含絕對值的式子化簡時的討論;
(2)此問重點考查了利用圖形抓住題目中的信息,分類討論的思想,還考查了圓錐曲線中的焦半徑公式(用點的一個坐標表示),還考查了兩點間的距離公式.