【答案】
分析:(1)本題需先根據(jù)已知條件,過(guò)C點(diǎn),設(shè)出該拋物線的解析式為y=ax
2+bx-2,再根據(jù)過(guò)A,B兩點(diǎn),即可得出結(jié)果.
(2)本題首先判斷出存在,首先設(shè)出橫坐標(biāo)和縱坐標(biāo),從而得出PA的解析式,再分三種情況進(jìn)行討論,當(dāng)
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時(shí)和
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時(shí),當(dāng)P,C重合時(shí),△APM≌△ACO,分別求出點(diǎn)P的坐標(biāo)即可.
(3)本題需先根據(jù)題意設(shè)出D點(diǎn)的橫坐標(biāo)和D點(diǎn)的縱坐標(biāo),再過(guò)D作y軸的平行線交AC于E,再由題意可求得直線AC的解析式為,即可求出E點(diǎn)的坐標(biāo),從而得出結(jié)果即可.
解答:解:(1)∵該拋物線過(guò)點(diǎn)C(0,-2),
∴可設(shè)該拋物線的解析式為y=ax
2+bx-2.
將A(4,0),B(1,0)代入,
得
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,
解得
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,
∴此拋物線的解析式為
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;
(2)存在.如圖,設(shè)P點(diǎn)的橫坐標(biāo)為m,
則P點(diǎn)的縱坐標(biāo)為
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,
當(dāng)1<m<4時(shí),AM=4-m,
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.
又∵∠COA=∠PMA=90°,
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∴①當(dāng)
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,
∵C在拋物線上,
∴OC=2,
∵OA=4,
∴
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,
∴△APM∽△ACO,
即
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.
解得m
1=2,m
2=4(舍去),
∴P(2,1).
②當(dāng)
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時(shí),△APM∽△CAO,即
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.
解得m
1=4,m
2=5(均不合題意,舍去)
∴當(dāng)1<m<4時(shí),P(2,1),
當(dāng)m>4時(shí),AM=m-4,PM=
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m
2-
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m+2,
①
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=
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=
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或②
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=
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=2,
把P(m,-
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m
2+
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m-2)代入得:2(
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m
2-
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m+2)=m-4,2(m-4)=
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m
2-
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m+2,
解得:第一個(gè)方程的解是m=-2-2
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<4(舍去)m=-2+2
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<4(舍去),
第二個(gè)方程的解是m=5,m=4(舍去)
求出m=5,-
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m
2+
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m-2=-2,
則P(5,-2),
當(dāng)m<1時(shí),AM=4-m,PM=
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m
2-
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m+2.
①
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=
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=
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或
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=
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=2,
則:2(
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m
2-
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m+2)=4-m,2(4-m)=
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m
2-
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m+2,
解得:第一個(gè)方程的解是m=0(舍去),m=4(舍去),第二個(gè)方程的解是m=4(舍去),m=-3,
m=-3時(shí),-
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m
2+
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m-2=-14,
則P(-3,-14),
綜上所述,符合條件的點(diǎn)P為(2,1)或(5,-2)或(-3,-14),
(3)如圖,設(shè)D點(diǎn)的橫坐標(biāo)為t(0<t<4),則D點(diǎn)的縱坐標(biāo)為|
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|.
過(guò)D作y軸的平行線交AC于E.
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由題意可求得直線AC的解析式為
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.
∴E點(diǎn)的坐標(biāo)為
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.
∴
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,
∴S
△DAC=S
△DCE+S
△DEA=
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DE•h+
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DE•(4-h)=
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DE•4,
∴
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,
∴當(dāng)t=2時(shí),△DAC面積最大,
∴D(2,1).
點(diǎn)評(píng):本題主要考查了二次函數(shù)解析式的確定、函數(shù)圖象交點(diǎn)的求法等知識(shí)點(diǎn),主要考查學(xué)生數(shù)形結(jié)合的數(shù)學(xué)思想方法.