【答案】
分析:(1)拋物線與x軸的交點(diǎn),即當(dāng)y=0,C點(diǎn)坐標(biāo)即當(dāng)x=0,分別令y以及x為0求出A,B,C坐標(biāo)的值;
(2)四邊形ACBP的面積=△ABC+△ABP,由A,B,C三點(diǎn)的坐標(biāo),可知△ABC是直角三角形,且AC=BC,則可求出△ABC的面積,根據(jù)已知可求出P點(diǎn)坐標(biāo),可知AP的長(zhǎng)度,以及點(diǎn)B到直線的距離,從而求出△ABP的面積,則就求出四邊形ACBP的面積;
(3)假設(shè)存在這樣的點(diǎn)M,兩個(gè)三角形相似,根據(jù)題意以及上兩題可知,∠PAC∠和∠MGA是直角,只需證明
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或
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即可.設(shè)M點(diǎn)坐標(biāo),根據(jù)題中所給條件可求出線段AG,CA,MG,CA的長(zhǎng)度,然后列等式,分情況討論,求解.
解答:解:(1)令y=0,
得x
2-1=0
解得x=±1,
令x=0,得y=-1
∴A(-1,0),B(1,0),C(0,-1);(2分)
(2)∵OA=OB=OC=1,
∴∠BAC=∠ACO=∠BCO=45°.
∵AP∥CB,
∴∠PAB=45°.
過點(diǎn)P作PE⊥x軸于E,則△APE為等腰直角三角形,
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令OE=a,則PE=a+1,
∴P(a,a+1).
∵點(diǎn)P在拋物線y=x
2-1上,
∴a+1=a
2-1.
解得a
1=2,a
2=-1(不合題意,舍去).
∴PE=3(4分).
∴四邊形ACBP的面積S=
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AB•OC+
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AB•PE
=
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×2×1+
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×2×3=4;(6分)
(3)假設(shè)存在
∵∠PAB=∠BAC=45°,
∴PA⊥AC
∵M(jìn)G⊥x軸于點(diǎn)G,
∴∠MGA=∠PAC=90°
在Rt△AOC中,OA=OC=1,
∴AC=
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在Rt△PAE中,AE=PE=3,
∴AP=3
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(7分)
設(shè)M點(diǎn)的橫坐標(biāo)為m,則M(m,m
2-1)
①點(diǎn)M在y軸左側(cè)時(shí),則m<-1.
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(�。┊�(dāng)△AMG∽△PCA時(shí),有
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.
∵AG=-m-1,MG=m
2-1.
即
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解得m
1=-1(舍去)m
2=
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(舍去).
(ⅱ)當(dāng)△MAG∽△PCA時(shí)有
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,
即
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.
解得:m=-1(舍去)m
2=-2.
∴M(-2,3)(10分).
②點(diǎn)M在y軸右側(cè)時(shí),則m>1
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(�。┊�(dāng)△AMG∽△PCA時(shí)有
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∵AG=m+1,MG=m
2-1
∴
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解得m
1=-1(舍去)m
2=
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.
∴M(
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,
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).
(ⅱ)當(dāng)△MAG∽△PCA時(shí)有
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,
即
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.
解得:m
1=-1(舍去)m
2=4,
∴M(4,15).
∴存在點(diǎn)M,使以A、M、G三點(diǎn)為頂點(diǎn)的三角形與△PCA相似
M點(diǎn)的坐標(biāo)為(-2,3),(
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,
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),(4,15).(13分)
點(diǎn)評(píng):考查拋物線與數(shù)軸交點(diǎn)求解問題,以及拋物線與三角形,四邊形之間關(guān)系轉(zhuǎn)換問題,相似三角形問題,要特別注意在第三問時(shí)要分情況討論.