【答案】
分析:(1)根據(jù)圓和拋物線的對(duì)稱性可知:點(diǎn)P必在拋物線的對(duì)稱軸上,根據(jù)B、C的坐標(biāo)可求出拋物線對(duì)稱軸的解析式即可得出圓P的半徑,連接PB,設(shè)拋物線對(duì)稱軸與x軸交于Q,那么PQ⊥x軸,且PQ=OA,已知了圓的半徑和BC的長(zhǎng),即可在直角三角形PBQ中求出PQ即OA的長(zhǎng),也就得出了A點(diǎn)坐標(biāo);
(2)將A、B、C三點(diǎn)坐標(biāo)代入拋物線中即可求出二次函數(shù)的解析式;
(3)先求出三角形AOB的面積,再根據(jù)題中給出的兩三角形的面積比得出三角形BCN的面積,BC長(zhǎng)為定值,可求出N點(diǎn)縱坐標(biāo)的絕對(duì)值,將其代入拋物線的解析式中即可求出N點(diǎn)的坐標(biāo);(由于N是直線BM與拋物線的交點(diǎn),且M在y軸負(fù)半軸,因此N點(diǎn)必在第一象限,據(jù)此可將不符合條件的N點(diǎn)坐標(biāo)舍去)
(4)根據(jù)弦切角定理可知:∠OAB=∠ADB,因此本題可分兩種情況:
①∠ABD=∠AOB=90°時(shí),此時(shí)MD⊥AB,且AD是圓P的直徑,可根據(jù)相似三角形AMB和DMA得出的關(guān)于MA、AD、AB、BD的對(duì)應(yīng)成比例線段求出MA的長(zhǎng),然后根據(jù)切割線定理可得出MB•MD=MA2,即可得出所求的值.
②∠BAD=∠AOB=90°時(shí),思路同①也是先求出MA的長(zhǎng),可根據(jù)直線MB的解析式求出M點(diǎn)坐標(biāo),然后通過(guò)相似三角形MAB和MDA(一個(gè)公共角,∠MBA和∠DAO都是90°加上一個(gè)等角)求出MA的長(zhǎng).后面同①.
解答:解:(1)A點(diǎn)坐標(biāo)是(0,2),⊙P的半徑長(zhǎng)為
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;
(2)拋物線的解析式是:y=
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x
2-
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x+2;
(3)設(shè)N點(diǎn)坐標(biāo)為(x
,y
),
由題意有
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BC•|y
|=
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OA•OB×
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∴
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×3y
=
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×2×1×
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解得y
=5
∵N點(diǎn)在拋物線上
∴
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x
2-
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x
+2=5
解得x
=6或x
=-1(不合題意,舍去)
∴N點(diǎn)的坐標(biāo)為(6,5);
(4)根據(jù)題意∠OAB=∠ADB,
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所以△AOB和△ABD相似有兩種情況
①∠ABD和∠AOB對(duì)應(yīng),此時(shí)AD是⊙P的直徑
則AB=
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,AD=5
∴BD=2
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∵Rt△AMB∽R(shí)t△DAB
∴MA:AD=AB:BD即MA=
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∵Rt△AMB∽R(shí)t△DMA
∴MA:MD=MB:MA
即MB•MD=MA
2=
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.
②∠BAD和∠AOB對(duì)應(yīng),此時(shí)BD是⊙P的直徑,
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所以直線MB過(guò)P點(diǎn)
∵B(1,0),P(
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,2)
∴直線MB的解析式是:y=
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x-
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∴M點(diǎn)的坐標(biāo)為(0,-
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)
∴AM=
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由△MAB∽△MDA得MA:MD=MB:MA
∴MB•MD=MA
2=
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.
點(diǎn)評(píng):本題考查了二次函數(shù)解析式的確定、圖形面積的求法、圓周角定理、切線的性質(zhì)、相似三角形的判定和性質(zhì)等知識(shí)點(diǎn).(4)題中,要根據(jù)相似三角形對(duì)應(yīng)邊和對(duì)應(yīng)角的不同分類討論,不要漏解.