試卷第1頁，共5頁15解：(1)因?yàn)閎sin2A＝asinB，所以2bsinAcosA＝asinB．根據(jù)正弦定理，得2sinBsinAcosA＝sinAsinB，………………2分因?yàn)閟inB≠0，sinA≠0，所以．…4分又，所以…6分（2）在△ABC中，由已知分由余弦定理可得a2=b2+c22bccosA，即分（1）證明：因?yàn)镻A是一條母線，所以PA丄平面ABC，又PA,ACC平面PAC且PAAC=A，所以BC丄平面PAC，而BCC平面PBC，則平面PAC丄平面PBC.·······················6分試卷第2頁，共5頁（2）設(shè)PA=2AB=4，則，因?yàn)镃是的中點(diǎn)，O為底面圓心，所以CO丄平面PAB，作OE丄PB，交PB于點(diǎn)E連接CE，則PB.OE=PA.BO，即在直角ΔCOE中解得.··························································································6分(2)比賽結(jié)束后，甲、乙個(gè)人得分可能為0，2，3，5.Ci,Di相互獨(dú)立，記兩輪投籃后甲總得分不低于8分為事件E，則E=C3D5+C5D3+C5D5，且C3D5，C5D3，C5D5彼此互斥.易得，························································································8分試卷第3頁，共5頁18解1）法1待定系數(shù)法）2，22①直線x2y+5=0與圓M相切，得到③,·······································6分法2幾何性質(zhì)）又因?yàn)閳A心M在直線2x+y—5=0上，聯(lián)立方程解得分設(shè)兩直線的交點(diǎn)為B(1,3)，由圓的幾何性質(zhì)，點(diǎn)B(1,3)在圓上，且為直線與圓的切點(diǎn)，又因?yàn)閳AM過點(diǎn)A(3,3)，且所以圓心M在直線x=2上，又圓心M也在直線2x+y—5=0上，聯(lián)立方程解得，故圓心M(2,1)，····································6分（2）設(shè)A(x,y)，因?yàn)锳為線段BD的中點(diǎn)，所以B(2x,2y+2)，因?yàn)锳，B在圓M上，所以解得或分試卷第4頁，共5頁19解1）由于橢圓的離心率為，故，又A1F222c2所以橢圓C的方程為.······································································4分（2i）設(shè)l與x軸交點(diǎn)為D，由于直線l交橢圓C于P、Q兩點(diǎn)（P、Q在x軸的兩側(cè)故直線l的的斜率不為0，直線l的方程為x=my+t，設(shè)P(x1,y1)，Q(x2,y2)，則分又直線l交與x軸不垂直可得k2+