题目内容
(2003•黄石)先阅读下面一段材料,再完成后面的问题:材料:过抛物线y=ax2(a>0)的对称轴上一点(0,-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_ST/2.png)
问题:若直线y=kx+b交抛物线y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_ST/3.png)
①求抛物线y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_ST/4.png)
②求证:直线AB过焦点时,CF⊥DF;
③当直线AB过点(-1,0),且以线段AB为直径的圆与准线l相切时,求这条直线对应的函数解析式.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_ST/images5.png)
【答案】分析:①将a=
代入题中给出的焦点坐标公式中即可.
②根据焦点的概念可知:AC=AF,BF=BD,如果连接CF、DF,那么CF必平分角AFO(可用三角形全等证出).同理可求得DF平分∠BFO,由此可得证.
③可连接圆心与切点,设圆心为M,切点为N,那么MN就是梯形ACDB的中位线,因此MN=
(AC+BD)=
AB,根据焦点的定义知:AF=AC,BF=BD,因此AF+BF=AB,也就是说直线AB恰好过焦点F,那么可根据F的坐标(①已求得)和已知的点(-1,0)的坐标用待定系数法求出抛物线的解析式.
解答:①解:F(0,1)
②证明:∵AC=AF,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/images3.png)
∴∠ACF=∠AFC
又∵AC∥OF,
∴∠ACF=∠CFO,
∴CF平分∠AFO,同理DF平分∠BFO;
而∠AFO+∠BFO=180°
∴∠CFO+∠DFO=
(∠AFO+∠BFO)=90°;
∴CF⊥DF.
③解:设圆心为M,且与l的切点为N,连接MN;
∴MN=
AB![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/images6.png)
在直角梯形ACDB中,M是AB的中点.
∴MN=
(AC+BD),而AC=AF,BD=BF.
∴MN=
(AF+BF)
∴AF+BF=AB
∴AB过焦点F(0,1).
又AB过点(-1,0)
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/7.png)
解得![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/8.png)
∴AB对应的函数解析式为y=x+1.
点评:本题为阅读类题,解题的关键是弄清材料中各定义的含义,然后结合自己掌握的知识进行求解.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/0.png)
②根据焦点的概念可知:AC=AF,BF=BD,如果连接CF、DF,那么CF必平分角AFO(可用三角形全等证出).同理可求得DF平分∠BFO,由此可得证.
③可连接圆心与切点,设圆心为M,切点为N,那么MN就是梯形ACDB的中位线,因此MN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/2.png)
解答:①解:F(0,1)
②证明:∵AC=AF,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/images3.png)
∴∠ACF=∠AFC
又∵AC∥OF,
∴∠ACF=∠CFO,
∴CF平分∠AFO,同理DF平分∠BFO;
而∠AFO+∠BFO=180°
∴∠CFO+∠DFO=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/3.png)
∴CF⊥DF.
③解:设圆心为M,且与l的切点为N,连接MN;
∴MN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/images6.png)
在直角梯形ACDB中,M是AB的中点.
∴MN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/5.png)
∴MN=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/6.png)
∴AF+BF=AB
∴AB过焦点F(0,1).
又AB过点(-1,0)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/7.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105614071142233/SYS201310191056140711422024_DA/8.png)
∴AB对应的函数解析式为y=x+1.
点评:本题为阅读类题,解题的关键是弄清材料中各定义的含义,然后结合自己掌握的知识进行求解.
![](http://thumb2018.1010pic.com/images/loading.gif)
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