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曲線C
1極坐標(biāo)方程為
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,即ρ=2sinθ,ρ
2=2ρsinθ
化為直角坐標(biāo)方程為x
2+y
2-2y=0.即x2+(y-1)2=1.
表示以C(0,1)為圓心,半徑為1 的圓.
C
2的極坐標(biāo)方程為,
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,即
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ρ(
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cosθ+
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sinθ)+1=0,
化為普通方程為x+y+1=0,表示一條直線
如圖,圓心到直線距離d=|CQ|=
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,曲線C
1上的點(diǎn)與曲線C
2上的點(diǎn)的最遠(yuǎn)距離為|PQ|=d+r=
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+1
(2)對(duì)于正實(shí)數(shù)x,y,由于a=
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,c=x+y≥2
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,b=p
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,且三角形任意兩邊之和大于第三邊,所以
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+2
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>b=p
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,且p
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+
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>2
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,p
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+2
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>
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,.
解得 1<p<3,故實(shí)數(shù)p的取值范圍是(1,3),
故答案為:
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+1,(1,3).
分析:(1)先將曲線的極坐標(biāo)方程方程化為普通方程,曲線C1的普通方程為x2+y2=2y,即x2+(y-1)2=1.表示以C(0,1)為圓心,半徑為1 的圓.曲線C2的普通方程為x+y+1=0,表示一條直線.利用直線和圓的位置關(guān)系求解.
(2)由基本不等式可得a≥
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,c≥2
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,再由三角形任意兩邊之和大于第三邊可得,
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+2
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>b=p
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,且p
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+
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>2
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,p
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+2
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>
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,由此求得實(shí)數(shù)p的取值范圍.
點(diǎn)評(píng):(1)本題以曲線參數(shù)方程出發(fā),考查了極坐標(biāo)方程、普通方程間的互化,直線和圓的位置關(guān)系.(2)本題主要考查基本不等式的應(yīng)用,注意不等式的使用條件,以及三角形中任意兩邊之和大于第三邊,屬于中檔題.