(1)设椭圆的方程为x2y219a2?b2?1(a?b?0)???3?b??a2?4,b2?1x2y2?3?椭圆的方程为?a2?b2?1334?1?1..............5分332)易得椭圆的方程为x2y2(2?1?1当直线的斜率不存在时，x?1,不合题意当直线的斜率存在时，直线设为y?k(x?1)??y?k(x?1)?x2y2得（2k2?1)x2?4k2x?2(k2?1)?0??2?1?1设P（x1,y1),Q(x2,y2)x4k22(k2?1)1?x2?2k2?1,x1x2?2k2?1F1P?(x1?1,y1),F1Q?(x2?1,y2)F1P?F1Q?0?(x1?1,y1)?(x2?1,y2)?x1x2?(x1?x2)?1?k2(x1?1)(x2?1)?(k2?1)x1x2?(k2?1)(x1?x2)?k2?1?7k2?1.2k2?1?0?k2?17k??77?直线方程为x?7y?1?0................................................................12分
6
20.(1)f/(x)?1a14?x2?x?f(x)在点（1，f(1))处的切线垂直于直线y?12x?f/(1)??34?a??2?a?54..................................................5分(2)由（1）知f(x)?x534?4x?lnx?2?f/?x2?4x?54x2（x?0)?f/(x)?0,x??1(舍去）或x?5f/(x)?0,0?x?5,f(x)在（0,5）内单调递减f/(x)?0,x?5,f(x)在（5,??）内单调递增x?5时，函数有极小值f(5)??ln5........................................12分
??21.(1)b?6.5,a?17.5?线性回归方程是y?6.5x?17.5.............................................9分(2)x?10代入（1）式得82.5万元..................................................3分22。(1)?=1,?=4???x/?x??y/?4y?x/?2 (2)..?=2,?=1?x2...?/1..........................................10分??y?2y
7
内蒙古集宁一中高二数学下学期第一次月考试题 文
