A,B是拋物線y2=2px(p>0)上的兩點,且OA⊥OB.
(1)求A,B兩點的橫坐標(biāo)之積和縱坐標(biāo)之積;
(2)求弦AB中點P的軌跡方程;
(3)求△AOB面積的最小值.
【答案】
分析:(1)先設(shè)出A,B,中點P的坐標(biāo),分別表示出AO,OB的斜率,利用二者垂直判斷出二者斜率乘積為-1求得x
1x
2+y
1y
2=0把拋物線的方程代入即可求得x
1x
2和y
1y
2.
(2)設(shè)出AO的方程代入拋物線求得x的值,進而表示出A的坐標(biāo),同理可表示出B的坐標(biāo),進而可表示出x
和y
,消去k即可求得二者的關(guān)系式,進而求得AB中點P的軌跡方程;
(3)根據(jù)S
△AOB=S
△AOM+S
△BOM,表示出△AOB面積,利用基本不等式求得面積的最小值.
解答:解:設(shè)A(x
1,y
1),B(x
2,y
2),中點P(x
,y
),
(1)k
0A=
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,k
OB=
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,
∵OA⊥OB,
∴x
1x
2+y
1y
2=0,
∵y
12=2px
1,y
22=2px
2,
∴
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•
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+y
1y
2=0
∴y
1y
2=-4p
2,x
1x
2=4p
2,
(2)設(shè)OA:y=kx,代入y
2=2px得x=0,x=
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,
∴A(
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,
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),同理以-
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代k得B(2pk
2,-2pk)
∴
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,消去k求得
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=(
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)
2+2,即y
2=px
-2p
2,即中點P軌跡方程為y
2=px-2p
2.
(3)S
△AOB=S
△AOM+S
△BOM=
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|OM|(|y
1|+|y
2|)=p(|y
1|+|y
2|)≥2p
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=4p
2當(dāng)且僅當(dāng)|y
1|=|y
2|時,等號成立
點評:本題主要考查了直線與圓錐曲線的綜合問題.解題的關(guān)鍵是靈活利用韋達定理,直線方程和曲線的方程聯(lián)立等.