七二五题
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/06/28 17:39:26
点A为抛物线C1:Y=½X²-2 的顶点,点B的坐标为(1,0),直线AB交抛物线C1于另一点C (1)求点C的坐标; (2)平行于y轴的直线x=3交直线AB于点D,交抛物线C1于点E,平行于y轴的直线x=a交直线AB于F,并抛物线C1于G,若FG:DE=4:3,求a的值; (3)将抛物线C1向下平移m(m>0)个单位得到抛物线C2,且抛物线C2的顶点为点p,交x轴负半轴于点M,交射线BC于点N,NQ垂直x轴于点Q,当NP平分角MNQ时,求m的值。
![七二五题](/uploads/image/z/4648640-32-0.jpg?t=%E4%B8%83%E4%BA%8C%E4%BA%94%E9%A2%98)
解题思路: 利用二次函数的性质解出
解题过程:
解:(1)当x=0时,y=﹣2;∴A(0,﹣2).
设直线AB的解析式为y=kx+b,则:
,解得
∴直线AB解析式为y=2x﹣2.
∵点C为直线y=2x﹣2与抛物线y=
x2﹣2的交点,则点C的横、纵坐标满足:
,解得
、
(舍)
∴点C的坐标为(4,6).
(2)直线x=3分别交直线AB和抛物线C1于D、E两点.
∴yD=4,yE=
,∴DE=
.
∵FG=DE=4:3,∴FG=2.
∵直线x=a分别交直线AB和抛物线C1于F、G两点.
∴yF=2a﹣2,yG=
a2﹣2
∴FG=|2a﹣
a2|=2,
解得:a1=2,a2=﹣2+2
,a3=2﹣2
.
(3)设直线MN交y轴于T,过点N做NH⊥y轴于点H;
设点M的坐标为(t,0),抛物线C2的解析式为y=
x2﹣2﹣m;
∴0=﹣
t2﹣2﹣m,∴﹣2﹣m=﹣
t2.
∴y=
x2﹣
t2,∴点P坐标为(0,﹣
t2).
∵点N是直线AB与抛物线y=
x2﹣
t2的交点,则点N的横、纵坐标满足:
,解得
、
(舍)
∴N(2﹣t,2﹣2t).
NQ=2﹣2t,MQ=2﹣2t,
∴MQ=NQ,∴∠MNQ=45°.
∴△MOT、△NHT均为等腰直角三角形,
∴MO=OT,HT=HN
∴OT=﹣t,NT=
(2﹣t),PT=﹣t+
t2.
∵PN平分∠MNQ,
∴PT=NT,
∴﹣t+
t2=
(2﹣t),
∴t1=﹣2
,t2=2(舍)
﹣2﹣m=﹣
t2=﹣
(﹣2
)2,∴m=2.
最终答案:略
解题过程:
解:(1)当x=0时,y=﹣2;∴A(0,﹣2).
设直线AB的解析式为y=kx+b,则:
![](http://img.wesiedu.com/upload/a/1d/a1df50df986654622475b3ca993af33b.png)
![](http://img.wesiedu.com/upload/f/f1/ff1a472e9f95f4bbc8159d9e6aae3c4b.png)
∴直线AB解析式为y=2x﹣2.
∵点C为直线y=2x﹣2与抛物线y=
![](http://img.wesiedu.com/upload/b/a7/ba7abfefc3de5a40a5ce940bd96d1fba.png)
![](http://img.wesiedu.com/upload/a/7a/a7af583c84faec99c9779d6008161304.png)
![](http://img.wesiedu.com/upload/8/6b/86b393c12694502c7c2e619573e9394e.png)
![](http://img.wesiedu.com/upload/2/f8/2f8227ca3b743091b78f9577fe6aedef.png)
∴点C的坐标为(4,6).
(2)直线x=3分别交直线AB和抛物线C1于D、E两点.
∴yD=4,yE=
![](http://img.wesiedu.com/upload/3/bb/3bb84da45601aafab5c630ae435cea5e.png)
![](http://img.wesiedu.com/upload/f/58/f58fa706b11b0aae8b032d7fde675e45.png)
∵FG=DE=4:3,∴FG=2.
∵直线x=a分别交直线AB和抛物线C1于F、G两点.
∴yF=2a﹣2,yG=
![](http://img.wesiedu.com/upload/0/ea/0ea25d867cd8688c267d94671148fb0c.png)
∴FG=|2a﹣
![](http://img.wesiedu.com/upload/0/ea/0ea25d867cd8688c267d94671148fb0c.png)
解得:a1=2,a2=﹣2+2
![](http://img.wesiedu.com/upload/9/07/907d2b60d234c0ffb1061f8afa1e5cd1.png)
![](http://img.wesiedu.com/upload/9/07/907d2b60d234c0ffb1061f8afa1e5cd1.png)
(3)设直线MN交y轴于T,过点N做NH⊥y轴于点H;
设点M的坐标为(t,0),抛物线C2的解析式为y=
![](http://img.wesiedu.com/upload/0/ea/0ea25d867cd8688c267d94671148fb0c.png)
∴0=﹣
![](http://img.wesiedu.com/upload/0/ea/0ea25d867cd8688c267d94671148fb0c.png)
![](http://img.wesiedu.com/upload/0/ea/0ea25d867cd8688c267d94671148fb0c.png)
∴y=
![](http://img.wesiedu.com/upload/0/ea/0ea25d867cd8688c267d94671148fb0c.png)
![](http://img.wesiedu.com/upload/9/29/929695d7b1489019c664d198f1077697.png)
![](http://img.wesiedu.com/upload/9/29/929695d7b1489019c664d198f1077697.png)
∵点N是直线AB与抛物线y=
![](http://img.wesiedu.com/upload/9/29/929695d7b1489019c664d198f1077697.png)
![](http://img.wesiedu.com/upload/9/29/929695d7b1489019c664d198f1077697.png)
![](http://img.wesiedu.com/upload/5/9f/59f96af2edfab9228a4da1c28d68e8d2.png)
![](http://img.wesiedu.com/upload/9/a6/9a69836088f08d4bb4af23c141d757ba.png)
![](http://img.wesiedu.com/upload/5/60/560acd2280aad01732731247b06721f9.png)
∴N(2﹣t,2﹣2t).
NQ=2﹣2t,MQ=2﹣2t,
∴MQ=NQ,∴∠MNQ=45°.
∴△MOT、△NHT均为等腰直角三角形,
∴MO=OT,HT=HN
∴OT=﹣t,NT=
![](http://img.wesiedu.com/upload/c/1a/c1a39176d55b6de7725bdc2913459247.png)
![](http://img.wesiedu.com/upload/9/ed/9ed654551aa4f92b35f91d8e8e172f58.png)
∵PN平分∠MNQ,
∴PT=NT,
∴﹣t+
![](http://img.wesiedu.com/upload/9/ed/9ed654551aa4f92b35f91d8e8e172f58.png)
![](http://img.wesiedu.com/upload/c/1a/c1a39176d55b6de7725bdc2913459247.png)
∴t1=﹣2
![](http://img.wesiedu.com/upload/c/1a/c1a39176d55b6de7725bdc2913459247.png)
﹣2﹣m=﹣
![](http://img.wesiedu.com/upload/9/ed/9ed654551aa4f92b35f91d8e8e172f58.png)
![](http://img.wesiedu.com/upload/9/ed/9ed654551aa4f92b35f91d8e8e172f58.png)
![](http://img.wesiedu.com/upload/c/1a/c1a39176d55b6de7725bdc2913459247.png)
![](http://img.wesiedu.com/upload/e/9d/e9d48adf5c8b5e1a69228bb6a283232a.png)
最终答案:略