题目内容
如图,抛物线C1:y=(x+m)2(m为常数,m>0),平移抛物线y=﹣x2,使其顶点D在抛物线C1位于y轴右侧的图象上,得到抛物线C2.抛物线C2交x轴于A,B两点(点A在点B的左侧),交y轴于点C,设点D的横坐标为a.
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(1)如图1,若m=
.
①当OC=2时,求抛物线C2的解析式;
②是否存在a,使得线段BC上有一点P,满足点B与点C到直线OP的距离之和最大且AP=BP?若存在,求出a的值;若不存在,请说明理由;
(2)如图2,当OB=2
﹣m(0<m<
)时,请直接写出到△ABD的三边所在直线的距离相等的所有点的坐标(用含m的式子表示).
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(1)如图1,若m=
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①当OC=2时,求抛物线C2的解析式;
②是否存在a,使得线段BC上有一点P,满足点B与点C到直线OP的距离之和最大且AP=BP?若存在,求出a的值;若不存在,请说明理由;
(2)如图2,当OB=2
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(1) ①y=﹣x2+
x+2.②
.(2)P1(
﹣m,1),P2(
﹣m,﹣3),P3(﹣
﹣m,3),P4(3
﹣m,3).
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试题分析:(1)①首先写出平移后抛物线C2的解析式(含有未知数a),然后利用点C(0,2)在C2上,求出抛物线C2的解析式;
②认真审题,题中条件“AP=BP”意味着点P在对称轴上,“点B与点C到直线OP的距离之和最大”意味着OP⊥BC.画出图形,如图1所示,利用三角函数(或相似),求出a的值;
(2)解题要点有3个:
i)判定△ABD为等边三角形;
ii)理论依据是角平分线的性质,即角平分线上的点到角两边的距离相等;
iii)满足条件的点有4个,即△ABD形内1个(内心),形外3个.不要漏解.
试题解析:(1)当m=
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∵抛物线C2的顶点D在抛物线C1上,且横坐标为a,
∴D(a,(a+
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∴抛物线C2:y=﹣(x﹣a)2+(a+
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①∵OC=2,∴C(0,2).
∵点C在抛物线C2上,
∴﹣(0﹣a)2+(a+
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解得:a=
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得抛物线C2的解析式为:y=﹣x2+
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②在(I)式中,
令y=0,即:﹣(x﹣a)2+(a+
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令x=0,得:y=a+
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设直线BC的解析式为y=kx+b,则有:
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∴直线BC的解析式为:y=﹣
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假设存在满足条件的a值.
∵AP=BP,
∴点P在AB的垂直平分线上,即点P在C2的对称轴上;
∵点B与点C到直线OP的距离之和≤BC,只有OP⊥BC时等号成立,
∴OP⊥BC.
如图1所示,设C2对称轴x=a(a>0)与BC交于点P,与x轴交于点E,
则OP⊥BC,OE=a.
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∵点P在直线BC上,
∴P(a,
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∵tan∠EOP=tan∠BCO=
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∴
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解得:a=
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∴存在a=
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(3)∵抛物线C2的顶点D在抛物线C1上,且横坐标为a,
∴D(a,(a+m)2).
∴抛物线C2:y=﹣(x﹣a)2+(a+m)2.
令y=0,即﹣(x﹣a)2+(a+m)2=0,解得:x1=2a+m,x2=﹣m,∴B(2a+m,0).
∵OB=2
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∴2a+m=2
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∴a=
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∴D(
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AB=OB+OA=2
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如图2所示,设对称轴与x轴交于点E,则DE=3,BE=AB=
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∵tan∠ABD=
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∴∠ABD=60°.
又∵AD=BD,∴△ABD为等边三角形.
作∠ABD的平分线,交DE于点P1,则P1E=BE•tan30°=
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∴P1(
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在△ABD形外,依次作各个外角的平分线,它们相交于点P2、P3、P4.
在Rt△BEP2中,P2E=BE•tan60°=
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∴P2(
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易知△ADP3、△BDP4均为等边三角形,∴DP3=DP4=AB=2
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∴P3(﹣
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综上所述,到△ABD的三边所在直线的距离相等的所有点有4个,
其坐标为:P1(
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【考点】二次函数综合题.
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