题目内容
:如图,在平面直角坐标系xoy中,抛物线y=x2-x-10与x轴的交点为A,与y轴的交点为点B,过点B作x轴的平行线BC,交抛物线于点C,连结AC.现有两动点P,Q分别从O,C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OC,PQ相交于点D,过点D作DE∥OA,交CA于点E,射线QE交x轴于点F.设动点P,Q移动的时间为t(单位:秒)
(1)求A,B,C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)求A,B,C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.
:略
:(1)在y=x2-x-10中,令y=0,得x2-8x-180=0.
解得x=-10或x=18,∴A(18,0).····················································· 1分
在y=x2-x-10中,令x=0,得y=-10.
∴B(0,-10).································· 2分
∵BC∥x轴,∴点C的纵坐标为-10.
由-10=x2-x-10得x=0或x=8.
∴C(8,-10).·································· 3分
∵y=x2-x-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QC∥PA,故只要QC=PA即可.
∵QC=t,PA=18-4t,∴t=18-4t.
解得t=.·························································································· 6分
(3)设点P运动了t秒,则OP=4t,QC=t,且0<t<4.5,说明点P在线段OA上,且不与点O,A重合.
∵QC∥OP, ∴====.
同理QC∥AF,∴===,即=.
∴AF=4t=OP.
∴PF=PA+AF=PA+OP=18.································································· 8分
∴S△PQF=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).
∴PQ2=(4t-8+t)2+102=(5t-8)2+100
FQ2=(18+4t-8+t)2+102=(5t+10)2+100.
①若FP=FQ,则182=(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QP=QF,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PQ=PF,则(5t-8)2+100=182.
即(5t-8)2=224,由于≈15,又0≤5t≤22.5,
∴-8≤5t-8≤14.5,而14.52=()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
解得x=-10或x=18,∴A(18,0).····················································· 1分
在y=x2-x-10中,令x=0,得y=-10.
∴B(0,-10).································· 2分
∵BC∥x轴,∴点C的纵坐标为-10.
由-10=x2-x-10得x=0或x=8.
∴C(8,-10).·································· 3分
∵y=x2-x-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QC∥PA,故只要QC=PA即可.
∵QC=t,PA=18-4t,∴t=18-4t.
解得t=.·························································································· 6分
(3)设点P运动了t秒,则OP=4t,QC=t,且0<t<4.5,说明点P在线段OA上,且不与点O,A重合.
∵QC∥OP, ∴====.
同理QC∥AF,∴===,即=.
∴AF=4t=OP.
∴PF=PA+AF=PA+OP=18.································································· 8分
∴S△PQF=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).
∴PQ2=(4t-8+t)2+102=(5t-8)2+100
FQ2=(18+4t-8+t)2+102=(5t+10)2+100.
①若FP=FQ,则182=(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QP=QF,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PQ=PF,则(5t-8)2+100=182.
即(5t-8)2=224,由于≈15,又0≤5t≤22.5,
∴-8≤5t-8≤14.5,而14.52=()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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