2016年5月7日福州高三质检文科数学含答案
················································· 7分 所以AP 2AF 2, ·
············································································ 8分 所以 ADP为正三角形， ·
····································································· 9分 所以AO
PD，且AO ·
因为AB 平面PAD，AB//CD，
所以CD 平面PAD． ·············································································· 10分 因为AO 平面PAD， 所以CD AO， ······················································································ 11分 又因为PD CD D，所以AO 平面PCD．
所以三棱锥A
PCD ································································ 12分 （20）本小题考查点与圆、直线与椭圆的位置关系等基础知识，考查运算求解能力、推理论证能力，考查数形结合思想、化归与转化思想等.满分12分．
解法一：
（Ⅰ）依题意得，2c a 4， ···················································· 2分 所以b2 a2 c2 1， ·················································································· 3分
x2
y2 1． ·所以E的方程为····································································· 4分 4
（Ⅱ）点A在 M外．理由如下： ································································ 5分 设P x1,y1 ,Q x2,y2 ，
y k(x 1),由 2得(1 4k2)x2 8k2x 4k2 4 0， ············································ 6分 2
x 4y 4,
所以， ( 8k2)2 4(4k2 1)(4k2 4) 48k2 16 0，
8k24k2 4
所以x1 x2 ，x1x2 . ····························································· 8分
1 4k21 4k2
因为AP x1 2,y1 ,AQ x2 2,y2 ，
所以AP AQ x1 2 x2 2 y1y2，
(1 k2)x1x2 (2 k2)(x1 x2) 4 k2
4(k2 1)(k2 1)8k2(2 k2)
········································ 10分 4 k2 ·22
1 4k1 4kk2
． 2
1 4k
因为k 0，所以AP AQ 0． 所以点A在 M外． ················································································ 12分 解法二：（Ⅰ）同解法一．
（Ⅱ）点A在 M外．理由如下： ································································ 5分 设P x1,y1 ,Q x2,y2 ，
y k(x 1),由 2得(1 4k2)x2 8k2x 4k2 4 0， ············································ 6分 2
x 4y 4,
所以， ( 8k2)2 4(4k2 1)(4k2 4) 48k2 16 0，
8k24k2 4
所以x1 x2 ，x1x2 . ····························································· 8分
1 4k21 4k2
