Question

：如圖，在平面直角坐標(biāo)系xoy中，拋物線y＝x²－x－10與x軸的交點(diǎn)為A，與y軸的交點(diǎn)為點(diǎn)B，過點(diǎn)B作x軸的平行線BC，交拋物線于點(diǎn)C，連結(jié)AC．現(xiàn)有兩動(dòng)點(diǎn)P，Q分別從O，C兩點(diǎn)同時(shí)出發(fā)，點(diǎn)P以每秒4個(gè)單位的速度沿OA向終點(diǎn)A移動(dòng)，點(diǎn)Q以每秒1個(gè)單位的速度沿CB向點(diǎn)B移動(dòng)，點(diǎn)P停止運(yùn)動(dòng)時(shí)，點(diǎn)Q也同時(shí)停止運(yùn)動(dòng)．線段OC，PQ相交于點(diǎn)D，過點(diǎn)D作DE∥OA，交CA于點(diǎn)E，射線QE交x軸于點(diǎn)F．設(shè)動(dòng)點(diǎn)P，Q移動(dòng)的時(shí)間為t(單位：秒)
（1）求A，B，C三點(diǎn)的坐標(biāo)和拋物線的頂點(diǎn)坐標(biāo)；
（2）當(dāng)t為何值時(shí)，四邊形PQCA為平行四邊形？請(qǐng)寫出計(jì)算過程；
（3）當(dāng)t∈（0，）時(shí)，△PQF的面積是否總為定值？若是，求出此定值；若不是，請(qǐng)說明理由；
（4）當(dāng)t為何值時(shí)，△PQF為等腰三角形？請(qǐng)寫出解答過程．

Answer 1

：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軸，∴點(diǎn)C的縱坐標(biāo)為－10．
由－10＝x²－x－10得x＝0或x＝8．
∴C(8，－10)．·································· 3分
∵y＝x²－x－10＝(x－4)²－
∴拋物線的頂點(diǎn)坐標(biāo)為(4，－)．    4分
（2）若四邊形PQCA為平行四邊形，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）設(shè)點(diǎn)P運(yùn)動(dòng)了t秒，則OP＝4t，QC＝t，且0＜t＜4.5，說明點(diǎn)P在線段OA上，且不與點(diǎn)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的面積總為定值90．································································· 9分
（4）設(shè)點(diǎn)P運(yùn)動(dòng)了t秒，則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，則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，則(5t－8)²＋100＝(5t＋10)²＋100．
即(5t－8)²＝(5t＋10)²，無0≤t≤4.5的t滿足．·································· 12分
③若PQ＝PF，則(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．
故無0≤t≤4.5的t滿足此方程．·························································· 13分
注：也可解出t＝＜0或t＝＞4.5均不合題意，
故無0≤t≤4.5的t滿足此方程．
綜上所述，當(dāng)t＝－2時(shí)，△PQF為等腰三角形．·························· 14分