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2023年全国乙卷理科数学解析几何大题的地阶作法与天阶解答

2023-07-12 16:52 作者:格心致力  | 我要投稿

高考数学中的解析几何大题、导数大题往往不仅仅只有一种方法可以攻克,一般而言至少有两种作法,下面我们就来看一下2023年全国乙卷理科数学解析几何大题的两种作法。

(2023 乙卷理科)解析几何:已知曲线C的方程为%5Cfrac%7By%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7Bx%5E2%7D%7Bb%5E2%7D%3D1%5Cleft(a%3Eb%3E0%5Cright),离心率为%5Cfrac%7B%5Csqrt5%7D%7B3%7D,曲线过点A%5Cleft(-2%2C0%5Cright)

⑴求曲线C的方程;


⑵过点%5Cleft(-2%2C1%5Cright)的直线交曲线C于P,Q两点,直线AP,AQ与y 轴交于M,N两点,证明:线段MN的中点是定点。


解:如图所示:⑴由离心率为%5Cfrac%7B%5Csqrt5%7D%7B3%7D得:

%5Cfrac%7Bc%7D%7Ba%7D%3D%5Cfrac%7B%5Csqrt5%7D%7B3%7D%20 

 

解析几何图形——格心原创

又由曲线过A%5Cleft(-2%2C0%5Cright)得:

%5Cfrac%7B0%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Cleft(-2%5Cright)%5E2%7D%7Bb%5E2%7D%3D1%20

再有 

c%5E2%2Bb%5E2%3Da%5E2%5C%20

联立①②③解得

a%3D3,b%3D2,c%3D%5Csqrt5

故曲线C的方程为

%5Cfrac%7By%5E2%7D%7B9%7D%2B%5C%20%5Cfrac%7Bx%5E2%7D%7B4%7D%3D1

⑵设P的坐标为%5Cleft(x_1%2Cy_1%5Cright),Q的坐标为%5Cleft(x_2%2Cy_2%5Cright) ,M的坐标为%5Cleft(0%2Cm%5Cright),N的坐标为%5Cleft(0%2Cn%5Cright)

MN的中点记为T%5Cleft(0%2Ct%5Cright)

 对A和P应用“截距坐标公式”得:


m%3D%5Cfrac%7B%5Cleft(-2%5Cright)%5Cast%20y_1-x_1%5Cast0%7D%7B%5Cleft(-2%5Cright)-x_1%7D%3D%5Cfrac%7B2y_1%7D%7Bx_%7B1%7D%2B2%7D

同理可得:

n%3D%5Cfrac%7B2y_2%7D%7Bx_%7B2%7D%2B2%7D

从而由“中点坐标公式”得:

t%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(m%2Bn%5Cright)%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(%5Cfrac%7B2y_1%7D%7Bx_%7B1%7D%2B2%7D%2B%5Cfrac%7B2y_1%7D%7Bx_%7B1%7D%2B2%7D%5Cright)%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%3D%20%5Cfrac%7By_2%7D%7Bx_%7B2%7D%2B2%7D%2B%5Cfrac%7By_2%7D%7Bx_%7B2%7D%2B2%7D%0A

                     

设过点%5Cleft(-2%2C3%5Cright)的直线方程为

y-3%3Dk%5Cleft(x-%5Cleft(-2%5Cright)%5Cright)

接下来的解题有以下两种选择:

方法一(地阶作法):

联立④⑤得:

%5Cleft(4k%5E2%2B9%5Cright)x%5E2%2B%5Cleft(16k%5E2%2B24k%5Cright)x%2B16k%5E2%2B48k%3D0

韦达定理得:

x_1%7B%2Bx%7D_2%3D-%5Cfrac%7B16k%5E2%2B24k%7D%7B4k%5E2%2B9%7D,x_1x_2%3D%5Cfrac%7B16k%5E2%2B48k%7D%7B4k%5E2%2B9%7D

于是

t%3D%5Cfrac%7Bk%5Cleft(x_1%2B2%5Cright)%2B3%7D%7Bx_1%2B2%7D%2B%5Cfrac%7Bk%5Cleft(x_2%2B2%5Cright)%2B3%7D%7Bx_2%2B2%7D%0A%3D2k%2B3%5Cfrac%7B%5Cleft(x_2%2B2%5Cright)%2B%5Cleft(x_1%2B2%5Cright)%7D%7B%5Cleft(x_2%2B2%5Cright)%5Cleft(x_1%2B2%5Cright)%7D%0A%3D2k%2B3%5Cfrac%7Bx_1%2Bx_2%2B4%7D%7Bx_1x_2%2B2%5Cleft(x_1%2Bx_2%5Cright)%2B4%7D%0A%0A

两根之和和两根之积带入得:

t%3D2k%2B3%5Cfrac%7B-%5Cfrac%7B16k%5E2%2B24k%7D%7B4k%5E2%2B9%7D%2B4%7D%7B%5Cfrac%7B16k%5E2%2B48k%7D%7B4k%5E2%2B9%7D-2%5Cfrac%7B16k%5E2%2B24k%7D%7B4k%5E2%2B9%7D%2B4%7D%0A%3D3%0A

因为t%3D3为定值

所以MN的中点为定点%5Cleft(0%2C3%5Cright)

从而命题得证。

方法二(天阶作法):

对④进行恒等变换

%5Cfrac%7By%5E2%7D%7B9%7D%2B%5C%20%5Cfrac%7B%5Cleft(x%2B2-2%5Cright)%5E2%7D%7B4%7D%3D1 

得:

%5Cfrac%7By%5E2%7D%7B9%7D%2B%5Cfrac%7B%5Cleft(x%2B2%5Cright)%5E2%7D%7B4%7D-(x%2B2)%3D0

对⑤进行恒等变换得:

                                              %5Cfrac%7By-k%5Cleft(x%2B2%5Cright)%7D%7B3%7D%3D1⑦                                            

将带入⑥中进行齐次处理得:

%5Cfrac%7By%5E2%7D%7B9%7D%2B%5Cfrac%7B%5Cleft(x%2B2%5Cright)%5E2%7D%7B4%7D-(x%2B2)%5Cleft(%5Cfrac%7By-k%5Cleft(x%2B2%5Cright)%7D%7B3%7D%5Cright)%3D0

对⑧进行化简整理得:

%5Cfrac%7By%5E2%7D%7B9%7D-%5Cfrac%7B1%7D%7B3%7D%5Cleft(x%2B2%5Cright)y%2B%5Cleft(%5Cfrac%7B1%7D%7B4%7D%2B%5Cfrac%7Bk%7D%7B3%7D%5Cright)%5Cleft(x%2B2%5Cright)%5E2%3D0

对⑨进行恒等变换得:

%5Cfrac%7B1%7D%7B9%7D%5Cleft(%5Cfrac%7By%7D%7Bx%2B2%7D%5Cright)%5E2-%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7By%7D%7Bx%2B2%7D%2B(%5Cfrac%7B1%7D%7B4%7D%2B%5Cfrac%7Bk%7D%7B3%7D)%3D0

韦达定理得:

%5Cfrac%7By_1%7D%7Bx_%7B1%7D%2B2%7D%2B%5Cfrac%7By_1%7D%7Bx_%7B1%7D%2B2%7D%3D-%5Cfrac%7B-%5Cfrac%7B1%7D%7B3%7D%7D%7B%5Cfrac%7B1%7D%7B9%7D%7D%3D3

可知

t%3D3

因而MN的中点为定点%5Cleft(0%2C3%5Cright)

从而命题得证。

评价与思考:此题的地阶作法经典常规,计算量较大,容易出错,是为多数同学熟知的方法;而天阶作法看起来步骤较多,其实其作法巧妙灵活,出人意料,技巧性强,但是对大部分同学来说并不了解或者精通其变换处理的技巧。要想彻底掌握这两种方法,还需要多加练习,总结,斟酌与思考!


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