【答案】
分析:(1)設(shè)拋物線的解析式為:y=a(x+2)
2+4,因?yàn)閽佄锞€過原點(diǎn),把(0,0)代入,求出a即可.
(2)由于PQ⊥MA,即∠MQP=∠MBA=90°;所以只要滿足∠PMQ=∠MAB或∠PMQ=∠AMB.
①∠PMQ=∠AMB時(shí),先找出點(diǎn)B關(guān)于直線MA的對(duì)稱點(diǎn)(設(shè)為點(diǎn)C),顯然有AC=AB=2、MC=MB=4,可根據(jù)該條件得到點(diǎn)C的坐標(biāo),進(jìn)而求出直線MC(即直線MP)的解析式,聯(lián)立拋物線的解析式即可得到點(diǎn)P的坐標(biāo);
②∠PMQ=∠MAB時(shí),若設(shè)直線MP與x軸的交點(diǎn)為D,那么△MAD必為等腰三角形,即MD=AD,根據(jù)此條件先求出點(diǎn)D的坐標(biāo),進(jìn)而得出直線MP的解析式,聯(lián)立拋物線的解析式即可得解.
解答:解:(1)∵過原點(diǎn)的拋物線的頂點(diǎn)為M(-2,4),
∴設(shè)拋物線的解析式為:y=a(x+2)
2+4,
將x=0,y=0代入可得:4a+4=0,
解得:a=-1,
∴拋物線解析式為:y=-(x+2)
2+4,
即y=-x
2-4x;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/images0.png)
(2)∵PQ⊥MA
∴∠MQP=∠MBA=90°;
若△MPQ、△MAB相似,那么需滿足下面的其中一種情況:
①∠PMQ=∠AMB,此時(shí)MA為∠PMB的角平分線,如圖①;
取點(diǎn)B關(guān)于直線MA的對(duì)稱點(diǎn)C,則AC=AB=2,MC=MB=4,設(shè)點(diǎn)C(x,y),有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/0.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/1.png)
(舍),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/2.png)
∴點(diǎn)C的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/4.png)
);
設(shè)直線MP的解析式:y=kx+b,代入M(-2,4)、(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/6.png)
)得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/7.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/8.png)
∴直線MP:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/9.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/10.png)
聯(lián)立拋物線的解析式,有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/11.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/13.png)
∴點(diǎn)P的坐標(biāo)(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/15.png)
);
②∠PMQ=∠MAB,如右圖②,此時(shí)△MAD為等腰三角形,且MD=AD,若設(shè)點(diǎn)D(x,0),則有:
(x+4)
2=(x+2)
2+(0-4)
2,解得:x=1
∴點(diǎn)D(1,0);
設(shè)直線MP的解析式:y=kx+b,代入M(-2,4)、D(1,0)后,有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/16.png)
,解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/17.png)
∴直線MP:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/18.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/19.png)
聯(lián)立拋物線的解析式有:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/20.png)
,解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/21.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/22.png)
∴點(diǎn)P的坐標(biāo)(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/24.png)
)
綜上,符合條件的P點(diǎn)有兩個(gè),且坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/26.png)
)、(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/27.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/28.png)
).
故答案:(1)y=-x
2-4x;(2)(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/29.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/30.png)
)、(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/31.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191709420805711/SYS201311011917094208057014_DA/32.png)
).
點(diǎn)評(píng):該題雖然是一道填空題,但難度不亞于壓軸題;主要的難度在于第二題,在“相似三角形→相等角→確定關(guān)鍵點(diǎn)→得到直線MP解析式”的解題思路中,綜合了相似三角形、等腰三角形的性質(zhì)、軸對(duì)稱圖形、坐標(biāo)系兩點(diǎn)間的距離公式、函數(shù)圖象交點(diǎn)坐標(biāo)的求法等重點(diǎn)知識(shí),這就要求同學(xué)們有扎實(shí)的基礎(chǔ)功底和良好的數(shù)形結(jié)合的思考方法.