题目内容

如图,在平面直角坐标系xoy中,抛物线yx 2x-10与x轴的交点为A,与y轴的交点为点B,过点Bx轴的平行线BC,交抛物线于点C,连结AC.现有两动点PQ分别从OC两点同时出发,点P以每秒4个单位的速度沿OA向终点A移动,点Q以每秒1个单位的速度沿CB向点B移动,点P停止运动时,点Q也同时停止运动.线段OCPQ相交于点D,过点DDEOA,交CA于点E,射线QEx轴于点F.设动点PQ移动的时间为t(单位:秒)

(1)求ABC三点的坐标和抛物线的顶点坐标;

(2)当t为何值时,四边形PQCA为平行四边形?请写出计算过程;

(3)当t∈(0)时,△PQF的面积是否总为定值?若是,求出此定值;若不是,请说明理由;

(4)当t为何值时,△PQF为等腰三角形?请写出解答过程.

 


(1)在yx 2x-10中,令y=0,得x 2-8x-180=0.

解得x=-10或x=18,∴A(18,0).········································ 1分

yx 2x-10中,令x=0,得y=-10.

B(0,-10).·························· 2分

BCx轴,∴点C的纵坐标为-10.

由-10=x 2x-10得x=0或x=8.

C(8,-10).························· 3分

yx 2x-10=(x-4)2

∴抛物线的顶点坐标为(4,-).············································· 4分

(2)若四边形PQCA为平行四边形,由于QCPA,故只要QCPA即可.

QCtPA=18-4t,∴t=18-4t

解得t.······································································· 6分

(3)设点P运动了t秒,则OP=4tQCt,且0<t<4.5,说明点P在线段OA上,且不与点OA重合.

QCOP,     ∴

同理QCAF,∴,即

AF=4tOP

PFPAAFPAOP=18.················································ 8分

SPQF PF·OB×18×10=90

∴△PQF的面积总为定值90.·················································· 9分

(4)设点P运动了t秒,则P(4t,0),F(18+4t,0),Q(8-t,-10)  t(0,4.5).

PQ 2=(4t-8+t)2+10 2=(5t-8)2+100

FQ 2=(18+4t-8+t)2+10 2=(5t+10)2+100.

①若FPFQ,则18 2=(5t+10)2+100.

即25(t+2)2=224,(t+2)2

∵0≤t≤4.5,∴2≤t+2≤6.5,∴t+2=

t-2.································································· 11分

②若QPQF,则(5t-8)2+100=(5t+10)2+100.

即(5t-8)2=(5t+10)2,无0≤t≤4.5的t满足.························· 12分

③若PQPF,则(5t-8)2+100=18 2

即(5t-8)2=224,由于≈15,又0≤5t≤22.5,

∴-8≤5t-8≤14.5,而14.5 2=()2<224.

故无0≤t≤4.5的t满足此方程.············································· 13分

注:也可解出t<0或t>4.5均不合题意,

故无0≤t≤4.5的t满足此方程.

综上所述,当t-2时,△PQF为等腰三角形.··················· 14分

 


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