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2023年全國乙卷理科數(shù)學(xué)解析幾何大題的地階作法與天階解答

2023-07-12 16:52 作者:格心致力  | 我要投稿

高考數(shù)學(xué)中的解析幾何大題、導(dǎo)數(shù)大題往往不僅僅只有一種方法可以攻克,一般而言至少有兩種作法,下面我們就來看一下2023年全國乙卷理科數(shù)學(xué)解析幾何大題的兩種作法。

(2023 乙卷理科)解析幾何:已知曲線C的方程為%5Cfrac%7By%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7Bx%5E2%7D%7Bb%5E2%7D%3D1%5Cleft(a%3Eb%3E0%5Cright),離心率為%5Cfrac%7B%5Csqrt5%7D%7B3%7D,曲線過點A%5Cleft(-2%2C0%5Cright)。

⑴求曲線C的方程;


⑵過點%5Cleft(-2%2C1%5Cright)的直線交曲線C于P,Q兩點,直線AP,AQ與y 軸交于M,N兩點,證明:線段MN的中點是定點。


解:如圖所示:⑴由離心率為%5Cfrac%7B%5Csqrt5%7D%7B3%7D得:

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

?

解析幾何圖形——格心原創(chuàng)

又由曲線過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

聯(lián)立①②③解得

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

故曲線C的方程為

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

⑵設(shè)P的坐標為%5Cleft(x_1%2Cy_1%5Cright),Q的坐標為%5Cleft(x_2%2Cy_2%5Cright)?,M的坐標為%5Cleft(0%2Cm%5Cright),N的坐標為%5Cleft(0%2Cn%5Cright)

MN的中點記為T%5Cleft(0%2Ct%5Cright)

?對A和P應(yīng)用“截距坐標公式”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

從而由“中點坐標公式”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

? ? ? ? ? ? ? ? ? ? ?

設(shè)過點%5Cleft(-2%2C3%5Cright)的直線方程為

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

接下來的解題有以下兩種選擇:

方法一(地階作法):

聯(lián)立④⑤得:

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

韋達定理得:

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

兩根之和和兩根之積帶入得:

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

因為t%3D3為定值

所以MN的中點為定點%5Cleft(0%2C3%5Cright)

從而命題得證。

方法二(天階作法):

對④進行恒等變換

%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

對⑤進行恒等變換得:

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

將帶入⑥中進行齊次處理得:

%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

對⑧進行化簡整理得:

%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

對⑨進行恒等變換得:

%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

韋達定理得:

%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的中點為定點%5Cleft(0%2C3%5Cright)

從而命題得證。

評價與思考:此題的地階作法經(jīng)典常規(guī),計算量較大,容易出錯,是為多數(shù)同學(xué)熟知的方法;而天階作法看起來步驟較多,其實其作法巧妙靈活,出人意料,技巧性強,但是對大部分同學(xué)來說并不了解或者精通其變換處理的技巧。要想徹底掌握這兩種方法,還需要多加練習(xí),總結(jié),斟酌與思考!


2023年全國乙卷理科數(shù)學(xué)解析幾何大題的地階作法與天階解答的評論 (共 條)

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