解:(1)設(shè)所求的二次函數(shù)的解析式為y=ax
2+bx+c,∵拋物線經(jīng)過A(4,-3),B(2,1)和C(-1,-8)三點(diǎn),
∴
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解之,得
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∴拋物線為y=-x
2+4x-3,令y=0,得-x
2+4x-3=0,解得x
1=1,x
2=3.
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∴拋物線與x軸的交點(diǎn)坐標(biāo)為M(1,0),N(3,0).
(2)過原點(diǎn)O作⊙G的切線,切點(diǎn)為D.易知OM=1,ON=3.由切割線定理,得OD
2=OM•ON=1×3.
∴OD=
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,即所求的切線OD長為
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.
(3)如右圖,連接DG,則∠ODG=90°,DG=1.∵OG=2,∴∠DOG=30°.
過D作DE⊥OG,垂足為E,則DE=OD•sin30°=
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,DE=OD•cos30°=
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.
∴點(diǎn)D的坐標(biāo)為D(
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,
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)或(
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,-
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).從而直線OD的解析式為y=±
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x.
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(4)Ⅰ、當(dāng)點(diǎn)P在x軸上方時(shí);
①點(diǎn)M是直角頂點(diǎn),此時(shí)MP1⊥x軸,即M、P1的橫坐標(biāo)相同;
當(dāng)x=1時(shí),y=
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x=
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;
即 P
1(1,
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);
②當(dāng)點(diǎn)P是直角頂點(diǎn)時(shí),由(2)知,P
2、D重合,即P
2(
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,
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);
③當(dāng)點(diǎn)N是直角頂點(diǎn),同①可求得 P
3(3,
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).
Ⅱ、當(dāng)點(diǎn)P在x軸下方時(shí),同Ⅰ可知:P
4(1,-
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),P
5(
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,-
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),P
6(3,-
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).
綜上,在直線OD上存在點(diǎn)P,使△MNP是直角三角形.所求P點(diǎn)的坐標(biāo)為(1,±
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),或(3,±
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),或(
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,±
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).
分析:(1)已知函數(shù)圖象上三個(gè)不同點(diǎn)的坐標(biāo),利用待定系數(shù)法即可求得拋物線的解析式;再令函數(shù)值為0,就能求出點(diǎn)M、N的坐標(biāo)(注意它們的位置).
(2)在(1)題中,已經(jīng)求得了M、N的坐標(biāo),則線段OM、ON的長可知,直接利用切割線定理即可求出OD的長.
(3)利用待定系數(shù)法求直線OD的解析式,必須先求出點(diǎn)D的坐標(biāo);連接圓心和切點(diǎn),過點(diǎn)D作x軸的垂線OE(垂足為E),首先由半徑長和OD的長求出∠DOG的度數(shù),然后在Rt△ODE中,通過解直角三角形求出DE、OE的長,則點(diǎn)D的坐標(biāo)可知,由此得解(需要注意的是:點(diǎn)D可能在x軸上方,也可能在x軸下方,所以直線OE的解析式應(yīng)該有兩個(gè)).
(4)在(3)中,已經(jīng)知道共有兩條直線OD,所以要分兩種大的情況討論,它們的解答方法是一致的,以點(diǎn)P在x軸上方為例進(jìn)行說明:
①當(dāng)點(diǎn)M是直角頂點(diǎn)時(shí),MP所在直線與x軸垂直,即M、P的橫坐標(biāo)相同,直接將點(diǎn)M的橫坐標(biāo)代入直線OD的解析式中即可得到點(diǎn)P的坐標(biāo);
②當(dāng)點(diǎn)P是直角頂點(diǎn)時(shí),由圓周角定理知:(2)題的切點(diǎn)D正好符合點(diǎn)P的條件;
③當(dāng)點(diǎn)N是直角頂點(diǎn)時(shí),方法同①.
點(diǎn)評(píng):此題是幾何與代數(shù)知識(shí)的綜合運(yùn)用,在考查常規(guī)知識(shí)的同時(shí),結(jié)合圓的對(duì)稱性等滲透了分類討論思想.解答(3)(4)問時(shí),解題者常拘泥于習(xí)慣性思維,只考慮到在x軸上方的切線OD和以P為直角頂點(diǎn)的Rt△MNP這些常見情形,從而導(dǎo)致丟解.作為壓軸題,本題(4)問顯示出了層次性,由易到難,逐步深入,體現(xiàn)了命題者的匠心.