摘要：22． P为椭圆上一点.左.右焦点分别为F1.F2. (1)若PF1的中点为M.求证|MO|=|PF1|, (2)若∠F1PF2=60°.求|PF­1||PF2|之值. (3)求|PF­1||PF2|的最值. 第Ⅰ卷选择题 1.▄ [B] [C] [D] 4.▄ [B] [C] [D] 7.▄ [B] [C] [D] 10.[A] ▄ [C] [D] 2.[A] ▄ [C] [D] 5.[A] [B] ▄ [D] 8.[A] [B] ▄ [D] 11.[A] [B] [C] ▄ 3.[A] [B] [C] ▄ 6.▄ [B] [C] [D] 9.[A] ▄[C] [D] 12.[A] [B] ▄ [D] ▄ 第Ⅱ卷 非选择题 二.填空题 (13) 相离 (14) (15) (0.5.2) (16) (-2.-1) 请在各题目的答题区域内作答.超出黑色矩形框限定区域的答案无效 三.解答题 (17) 证明:因为.分别为.的中点. 所以MN平行于 BD 又因为BD不在平面AMN内.MN在平面AMN内 所以面, (18)解: 解:设所求双曲线方程为: 双曲线过点M , 得双曲线方程为 (19) 解: (Ⅰ) 要使圆的面积最小.则AB为圆的直径. ∴所求圆的方程为(x-2)(x+2)+(y+3)(y+5)＝0.即 x2+(y+4)2＝5.····················································································· 5分 (Ⅱ) 因为kAB＝12.AB中点为. 所以AB中垂线方程为y+4＝-2x.即2x+y+4＝0.····························· 8分 解方程组得即圆心为. 根据两点间的距离公式.得半径r＝, 因此.所求的圆的方程为(x+1)2+(y+2)2＝10.····································· 12分 另解:设所求圆的方程为(x-a)2+(y-b)2＝r2.根据已知条件得 所以所求圆的方程为(x+1)2+(y+2)2＝10.

网址：http://m.1010jiajiao.com/timu_id_3308139[举报]