高考數(shù)學(xué)90個考點(diǎn)90個專題直線與圓錐曲線相交的綜合問題是每年高考的必考【解析】設(shè)P(x1,y1),Q(x2,y2)消去x得(16-9k2)x2-18kmx-9(m2+16)=0依題意可得kDA?kDB=-1∴(7m+75k)(m-3k)=0①當(dāng)7m+75k=0即m=-回代直線得y=kx-恒過定點(diǎn)②當(dāng)m-3k=0,即m=3k回代直線y=kx+3k,恒過定點(diǎn)(-3,0)不符合題意的點(diǎn)P(1,(到兩點(diǎn)的距離之和為4.(2)設(shè)M(x1,y1),N(x2,y2)，因為直線y=kx+m與橢圓C相交于M，N兩個不同的點(diǎn)，因為直線OM,ON的斜率之積等于-，又O到直線MN的距離為設(shè)P(x1,y1),Q(x2,y2)消去x得y2-8my-32=08m,y1y2=-32即kOP?kOQ=-2設(shè)P(x1,y1),Q(x2,y2)是直線PQ：mx+ny=1與拋物線y2=8x的兩個交點(diǎn)由1得y2-8x∴y2-8nxy-8mx2=02-8n-8m=0又直線mx+ny=1過點(diǎn)(4,0)∴kOP?kOQ=-8m=-2交x=-2于M,N兩點(diǎn)，求證:直線TM，TN的斜率之積為定值；設(shè)P(x1,y1),Q(x2,y2)消去x得y2-8my-32=08m,y1y2=-32聯(lián)立y-0=聯(lián)立y-0=即kTM?kTN=-(0,1),B2(0,-1),過點(diǎn)R(0,-)作與y軸不重合的直線l交橢圓解：由垂徑定理可知kPB1?kPB2=e2-1=-所以要證直線B2P與B1線B1P與B1Q斜率之積是為定值設(shè)直線l:y=kx-∴kB1P×kB1Q=-1?kPB2=e2-1=-【解析】解法1（設(shè)而不求逐步消元法設(shè)A(x1,y1),B(x2,y2)聯(lián)立8得y2-8my-64=0∴y1+y2=8m,y1y2=-64設(shè)圓過定點(diǎn)T(a,b)要使?m∈R都有kTA?kTB=-1恒成立即定點(diǎn)T為(0,0)2px(p>0)上有一點(diǎn)P(2,2)k2為定值.2px(p>0)過點(diǎn)P(2,2)，(2)依題意，過點(diǎn)M(2,0)的直線AB不垂直于y軸，設(shè)直線AB方程為x設(shè)A(x1,y1),B(x2,y2)，則y1+y2=2m，y1y2=-4，所以k1k2為定值-1.練2.若Rt△APB內(nèi)接于拋物線y2=8x，P(2,4)為直角頂點(diǎn),求證：斜邊AB過定點(diǎn)；設(shè)P(x1,y1),Q(x2,y2)消去x得y2-8my-8t=0∴y1+y2=8m,y1y2=-8t依題意kPA?kPB=-1即(y1-4)(y2-4)+(x1-2)(x2-2)=0∴y1y2-4(y1+y2)+16+(my1+t-2)(my2+t-2)=0∴y1y2-4(y1+y2)+16+m2y1y2+m(t-2)(y1+y2)+(t-2)2=0∴(1+m2)(-8t)+(-4+mt-2m)8m+16+(t-2)2=032m-(t-10)(t-2)=0因式分解為(4m-(t-10))(4m+t-2)=0∴4m-(t-10)=0或4m+t-2=0整理為m(y+4)+(10-x)=0可得定點(diǎn)(10,-4)②當(dāng)4m+t-2=0時t=2-4m回代直線得x=my+2-4m整理為m(y-4)+2-x=0可得定點(diǎn)綜上定點(diǎn)為(10,-4)解法2（設(shè)而不求齊次化設(shè)直線AB的方程為m(x-2)+n(y-4)=1將1替換為m(x-2)+n(y-4)得(8n+1)(y-4)2+(8m-8n)(x-2)(y-4)-8m(x-2)2=02得-8m=0由題意可知kPA×kPB=-1回代直線方程可得m(x-2)+n(y-4)=1(8n+1)(x-2)+8n(y-4)=8整理為8n(x+y-6)+x-10=0∴過定點(diǎn)(10,-4)設(shè)A(x1,y1),B(x2,y2)所在直線的方程為x=my+t∴y2-8my-8t=0依題意kPA?kPB=-1∴(x1-2)(x2-2)+(y1-4)(y2-4)=0∴(my1+t-2)(my2+t-2)+(y1-4)(y2-4)=0設(shè)y2-8my-8t=(y-y1)(y-y2)=(y1-y)(y-y2)=(t-2)2-16m2令y=4則(y1-4)(y2-4)=16-32m-8t∴(t-2)2-16m2+16-32m-8t=0得到(4m+t-2)(4m-t+10)=0∴4m+t-2=0或者4m-t+10=0當(dāng)4m+t-2=0時t=2-4m此時直線x=my+2-4m過定點(diǎn)(2,4)練3.已知橢圓C的中心是坐標(biāo)原點(diǎn)，直線x-2y-4經(jīng)過橢圓C的兩個頂點(diǎn).直線于M，N兩點(diǎn)，不妨設(shè)直線PQ：x