【答案】
分析:由題意可知AC=5,OA=3,根據(jù)勾股定理可知,OC=4,可知C點(diǎn)坐標(biāo),同理求出B點(diǎn)坐標(biāo),OA=3,AD=5,求出OD=2,求出D點(diǎn)坐標(biāo).
(1)∵點(diǎn)A的坐標(biāo)為(0,-3),線段AD=5,∴點(diǎn)D的坐標(biāo)(0,2).
連接AC,在Rt△AOC中,∠AOC=90°,OA=3,AC=5,∴OC=4.
∴點(diǎn)C的坐標(biāo)為(4,0);
同理可得點(diǎn)B坐標(biāo)為(-4,0).
(2)已知B,C,D三點(diǎn)坐標(biāo),設(shè)出解析式,代入即可求出函數(shù)解析式.
設(shè)所求二次函數(shù)的解析式為y=ax
2+bx+c,
由于該二次函數(shù)的圖象經(jīng)過B,C,D三點(diǎn),則
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解得
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∴所求的二次函數(shù)的解析式為y=-
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x
2+2;
(3)根據(jù)圖象可知,正切為
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,則∠cpf為直角,設(shè)出P點(diǎn)坐標(biāo),然后表示出CP,PF的長度,然后分情況討論
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=
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還是
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,或是兩者都可,求出P點(diǎn)坐標(biāo).
設(shè)點(diǎn)P坐標(biāo)為(t,0),由題意得t>5,
且點(diǎn)F的坐標(biāo)為(t,-
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t
2+2),PC=t-4,PF=
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t
2-2,
∵∠CPF=90°,∴當(dāng)△CPF中一個內(nèi)角的正切值為
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時,
①若
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時,即
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,解得t
1=12,t
2=4(舍);
②當(dāng)
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時,
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解得t
1=0(舍),t
2=4(舍),
所以所求點(diǎn)P的坐標(biāo)為(12,0).
解答:解:(1)∵點(diǎn)A的坐標(biāo)為(0,-3),線段AD=5,
∴點(diǎn)D的坐標(biāo)(0,2).(1分)
連接AC,在Rt△AOC中,∠AOC=90°,OA=3,AC=5,
∴OC=4.(1分)
∴點(diǎn)C的坐標(biāo)為(4,0);(1分)
同理可得點(diǎn)B坐標(biāo)為(-4,0).(1分)
(2)設(shè)所求二次函數(shù)的解析式為y=ax
2+bx+c,
由于該二次函數(shù)的圖象經(jīng)過B,C,D三點(diǎn),則
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(3分)
解得
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∴所求的二次函數(shù)的解析式為y=-
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x
2+2;(1分)
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(3)設(shè)點(diǎn)P坐標(biāo)為(t,0),由題意得t>5,(1分)
且點(diǎn)F的坐標(biāo)為(t,-
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t
2+2),PC=t-4,PF=
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t
2-2,
∵∠CPF=90°,
∴當(dāng)△CPF中一個內(nèi)角的正切值為
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時,
①若
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時,即
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,解得t
1=12,t
2=4(舍);(1分)
②當(dāng)
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時,
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解得t
1=0(舍),t
2=4(舍),(1分)
所以所求點(diǎn)P的坐標(biāo)為(12,0).(1分)
點(diǎn)評:本題旨在考查圓在坐標(biāo)中出現(xiàn)的問題,圓與拋物線交點(diǎn)問題,以及三角形中正切的概念.