過x軸上的動點A(a,0).引拋物線的兩條切線AP.AQ.P.Q為切點(1)若切線AP.AQ的斜率分別為k1.k2.求證k1?k2為定值;(2)求證:直線PQ過定點; 查看更多

 

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過x軸上的動點A(a,0)引拋物線y=x2+1的兩切線AP,AQ.P,Q為切點.
(I)求切線AP,AQ的方程;
(Ⅱ)求證直線PQ過定點;
(III)若a≠0,試求
S△APQ|OA|
的最小值.

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過x軸上的動點A(a,0)引拋物線y=x2+1的兩切線AP,AQ.P,Q為切點.
(I)求切線AP,AQ的方程;
(Ⅱ)求證直線PQ過定點;
(III)若a≠0,試求
S△APQ
|OA|
的最小值.

查看答案和解析>>

過x軸上的動點A(a,0)引拋物線y=x2+1的兩切線AP,AQ.P,Q為切點.
(I)求切線AP,AQ的方程;
(Ⅱ)求證直線PQ過定點;
(III)若a≠0,試求的最小值.

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過x軸上的動點A(a,0)的拋物線y=x2+1引兩切線AP、AQ,P、Q為切點.
(1)若切線AP,AQ的斜率分別為k1,k2,求證:k1•k2為定值;
(2)求證:直線PQ過定點;
(3)若a≠0,試求S△APQ:|OA|的最小值.

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過x軸上的動點A(a,0)的拋物線y=x2+1引兩切線AP、AQ,P、Q為切點.
(1)若切線AP,AQ的斜率分別為k1,k2,求證:k1k2為定值;
(2)求證:直線PQ過定點;
(3)若a≠0,試求S△APQ:|OA|的最小值.

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一.選擇題

1―5  CBABA   6―10  CADDA

二.填空題

11.       12.()       13.2          14.         15.

16.(1,4)

三.解答題

數(shù)學(xué)理數(shù)學(xué)理17,解:①         =2(1,0)                      (2分)             

        ?,                                        (4分)

?

        cos              =

 

        由,  ,    即B=              (6分)

                                               (7分)

                                                        (9分)

,                                                         (11分)

的取值范圍是(,1                                                      (13分)

18.解:①設(shè)雙曲線方程為:  ()                                 (1分)

由橢圓,求得兩焦點,                                           (3分)

,又為一條漸近線

, 解得:                                                     (5分)

                                                    (6分)

②設(shè),則                                                      (7分)

      

?                             (9分)

,  ?              (10分)

                                                (11分)

  ?

?                                        (13分)

        單減區(qū)間為[]        (6分)

       

      ②(i)當                                                      (8分)

      (ii)當,

      ,  (),,

      則有                                                                     (10分)

                                                     (11分)

        在(0,1]上單調(diào)遞減                     (12分)

                                                       (13分)

      20.解:①       

                                                              (2分)

      從而數(shù)列{}是首項為1,公差為C的等差數(shù)列

        即                                (4分)

       

         即………………※              (6分)

      當n=1時,由※得:c<0                                                    (7分)

      當n=2時,由※得:                                                 (8分)

      當n=3時,由※得:                                                 (9分)

          (

                                                (11分)

                               (12分)

      綜上分析可知,滿足條件的實數(shù)c不存在.                                    (13分)

      21.解:①設(shè)過A作拋物線的切線斜率為K,則切線方程:

                                                                      (2分)

          即

                                                                                                         (3分)

      ②設(shè)   又

           

                                                               (4分)

      同理可得 

                                                      (5分)

      又兩切點交于 

                                     (6分)

      ③由  可得:

       

                                                      (8分)

                        (9分)

       

       

       

                                                           (11分)

      當且僅當,取 “=”,此時

                                             (12分)

      22.①證明:由,    

        即證

        ()                                    (1分)

        

            即:                          (3分)

        ()    

         

         

                                                               (6分)

      ②由      

      數(shù)列

                                                    (8分)

      由①可知, 

                          (10分)

      由錯位相減法得:                                       (11分)

                                          (12分)

       

       


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