【答案】
分析:(1)已知了拋物線的對(duì)稱軸解析式,可用頂點(diǎn)式二次函數(shù)通式來設(shè)拋物線,然后將A、B兩點(diǎn)坐標(biāo)代入求解即可.
(2)平行四邊形的面積為三角形OEA面積的2倍,因此可根據(jù)E點(diǎn)的橫坐標(biāo),用拋物線的解析式求出E點(diǎn)的縱坐標(biāo),那么E點(diǎn)縱坐標(biāo)的絕對(duì)值即為△OAE的高,由此可根據(jù)三角形的面積公式得出△AOE的面積與x的函數(shù)關(guān)系式進(jìn)而可得出S與x的函數(shù)關(guān)系式.
①將S=24代入S,x的函數(shù)關(guān)系式中求出x的值,即可得出E點(diǎn)的坐標(biāo)和OE,OA的長;如果平行四邊形OEAF是菱形,則需滿足平行四邊形相鄰兩邊的長相等,據(jù)此可判斷出四邊形OEAF是否為菱形.
②如果四邊形OEAF是正方形,那么三角形OEA應(yīng)該是等腰直角三角形,即E點(diǎn)的坐標(biāo)為(3,-3)將其代入拋物線的解析式中即可判斷出是否存在符合條件的E點(diǎn).
解答:解:(1)因?yàn)閽佄锞€的對(duì)稱軸是x=
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,
設(shè)解析式為y=a(x-
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)
2+k.
把A,B兩點(diǎn)坐標(biāo)代入上式,得
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,
解得a=
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,k=-
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.
故拋物線解析式為y=
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(x-
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)
2-
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,頂點(diǎn)為(
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,-
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).
(2)∵點(diǎn)E(x,y)在拋物線上,位于第四象限,且坐標(biāo)適合y=
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(x-
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)
2-
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,
∴y<0,
即-y>0,-y表示點(diǎn)E到OA的距離.
∵OA是OEAF的對(duì)角線,
∴S=2S
△OAE=2×
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×OA•|y|=-6y=-4(x-
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)
2+25.
因?yàn)閽佄锞€與x軸的兩個(gè)交點(diǎn)是(1,0)和(6,0),
所以自變量x的取值范圍是1<x<6.
①根據(jù)題意,當(dāng)S=24時(shí),即-4(x-
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)
2+25=24.
化簡(jiǎn),得(x-
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)
2=
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.
解得x
1=3,x
2=4.
故所求的點(diǎn)E有兩個(gè),
分別為E
1(3,-4),E
2(4,-4),
點(diǎn)E
1(3,-4)滿足OE=AE,
所以平行四邊形OEAF是菱形;
點(diǎn)E
2(4,-4)不滿足OE=AE,
所以平行四邊形OEAF不是菱形;
②當(dāng)OA⊥EF,且OA=EF時(shí),平行四邊形OEAF是正方形,
此時(shí)點(diǎn)E的坐標(biāo)只能是(3,-3),
而坐標(biāo)為(3,-3)的點(diǎn)不在拋物線上,
故不存在這樣的點(diǎn)E,使平行四邊形OEAF為正方形.
點(diǎn)評(píng):本題主要考查了二次函數(shù)解析式的確定、圖形面積的求法、平行四邊形的性質(zhì)、菱形和正方形的判定等知識(shí).綜合性強(qiáng),難度適中.