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不想埋头傻算,那就来改造命题(2020北京卷圆锥曲线)

2022-07-12 21:58 作者:数学老顽童  | 我要投稿

(2020北京,20)已知椭圆C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1过点A%5Cleft(%20-2%2C-1%20%5Cright)%20,且a%3D2b.

(1)求椭圆C的方程;

(2)过点B%5Cleft(%20-4%2C0%20%5Cright)%20的直线l交椭圆C于点MN,直线MANA分别交直线x%3D-4于点PQ.求%5Cfrac%7B%5Cleft%7C%20PB%20%5Cright%7C%7D%7B%5Cleft%7C%20BQ%20%5Cright%7C%7D的值.

解:(1)由题可知%5Cfrac%7B4%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D%3D1

又因为a%3D2b

二者联立,解得a%3D2%5Csqrt%7B2%7D%20b%3D%5Csqrt%7B2%7D%20

所以椭圆C的方程为%5Cfrac%7Bx%5E2%7D%7B8%7D%2B%5Cfrac%7By%5E2%7D%7B2%7D%3D1.

(2)先画个图

先猜再证:%5Cfrac%7B%5Cleft%7C%20PB%20%5Cright%7C%7D%7B%5Cleft%7C%20BQ%20%5Cright%7C%7D%3D1

其等价于y_P%2By_Q%3D2y_B

若上式成立,则

%5Cbegin%7Baligned%7D%09k_%7BPA%7D%2Bk_%7BQA%7D%26%3D%5Cfrac%7By_P-y_A%7D%7Bx_P-x_A%7D%2B%5Cfrac%7By_Q-y_A%7D%7Bx_Q-x_A%7D%5C%5C%09%26%3D%5Cfrac%7By_P-y_A%7D%7Bx_B-x_A%7D%2B%5Cfrac%7By_Q-y_A%7D%7Bx_B-x_A%7D%5C%5C%09%26%3D%5Cfrac%7By_P%2By_Q-2y_A%7D%7Bx_B-x_A%7D%5C%5C%09%26%3D%5Cfrac%7B2y_B-2y_A%7D%7Bx_B-x_A%7D%5C%5C%09%26%3D2%5Ccdot%20%5Cfrac%7By_B-y_A%7D%7Bx_B-x_A%7D%5C%5C%09%26%3D2k_%7BBA%7D%3D-1%5C%5C%5Cend%7Baligned%7D

故只需证k_%7BPA%7D%2Bk_%7BQA%7D%3D-1

即证k_%7BMA%7D%2Bk_%7BNA%7D%3D-1.

下面开证(方法:齐次化联立)

先改写椭圆C的方程:

%5Cfrac%7B%5Cleft(%20x%2B2%20%5Cright)%20%5E2-4x-4%7D%7B8%7D%2B%5Cfrac%7B%5Cleft(%20y%2B1%20%5Cright)%20%5E2-2y-1%7D%7B2%7D%3D1

再整理一下:

%5Cfrac%7B%5Cleft(%20x%2B2%20%5Cright)%20%5E2%7D%7B8%7D%2B%5Cfrac%7B%5Cleft(%20y%2B1%20%5Cright)%20%5E2%7D%7B2%7D-%5Cleft%5B%20%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x%2B2%20%5Cright)%20%2B%5Cleft(%20y%2B1%20%5Cright)%20%5Cright%5D%20%3D0.

设直线l的方程为

m%5Cleft(%20x%2B2%20%5Cright)%20%2Bn%5Cleft(%20y%2B1%20%5Cright)%20%3D1

先保证其不过点A

因为l过点B,(再保证其必过点B

所以m%5Cleft(%20-4%2B2%20%5Cright)%20%2Bn%5Cleft(%200%2B1%20%5Cright)%20%3D1

整理得n%3D2m%2B1

所以l的方程为

m%5Cleft(%20x%2B2%20%5Cright)%20%2B%5Cleft(%202m%2B1%20%5Cright)%20%5Cleft(%20y%2B1%20%5Cright)%20%3D1.

联立椭圆C与直线l,得

%5Cfrac%7B%5Cleft(%20x%2B2%20%5Cright)%20%5E2%7D%7B8%7D%2B%5Cfrac%7B%5Cleft(%20y%2B1%20%5Cright)%20%5E2%7D%7B2%7D-%5Cleft%5B%20%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x%2B2%20%5Cright)%20%2B%5Cleft(%20y%2B1%20%5Cright)%20%5Cright%5D%20%5Cleft%5B%20m%5Cleft(%20x%2B2%20%5Cright)%20%2B%5Cleft(%202m%2B1%20%5Cright)%20%5Cleft(%20y%2B1%20%5Cright)%20%5Cright%5D%20%3D0

展开,

%5Cfrac%7B%5Cleft(%20x%2B2%20%5Cright)%20%5E2%7D%7B8%7D%2B%5Cfrac%7B%5Cleft(%20y%2B1%20%5Cright)%20%5E2%7D%7B2%7D-%5Cfrac%7B1%7D%7B2%7Dm%5Cleft(%20x%2B2%20%5Cright)%20%5E2-%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Cleft(%20x%2B2%20%5Cright)%20%5Cleft(%20y%2B1%20%5Cright)%20-%5Cleft(%202m%2B1%20%5Cright)%20%5Cleft(%20y%2B1%20%5Cright)%20%5E2%3D0

并项,

%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Cleft(%20y%2B1%20%5Cright)%20%5E2%2B%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Cleft(%20x%2B2%20%5Cright)%20%5Cleft(%20y%2B1%20%5Cright)%20%2B%5Cleft(%20%5Cfrac%7B1%7D%7B2%7Dm-%5Cfrac%7B1%7D%7B8%7D%20%5Cright)%20%5Cleft(%20x%2B2%20%5Cright)%20%5E2%3D0

各项同除以%5Cleft(%20x%2B2%20%5Cright)%20%5E2

%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Cleft(%20%5Cfrac%7By%2B1%7D%7Bx%2B2%7D%20%5Cright)%20%5E2%2B%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Ccdot%20%5Cfrac%7By%2B1%7D%7Bx%2B2%7D%2B%5Cfrac%7B1%7D%7B2%7Dm-%5Cfrac%7B1%7D%7B8%7D%3D0

再修饰一哈:

%5Cleft(%20%5Cfrac%7By%2B1%7D%7Bx%2B2%7D%20%5Cright)%20%5E2%2B%5Cfrac%7By%2B1%7D%7Bx%2B2%7D%2B%5Cfrac%7B1%7D%7B4%7D%5Ccdot%20%5Cfrac%7B4m-1%7D%7B4m%2B1%7D%3D0

竟然颇具美感!

显然

k_%7BAM%7D%2Bk_%7BAN%7D%3D%5Cfrac%7By_1%2B1%7D%7Bx_1%2B2%7D%2B%5Cfrac%7By_2%2B1%7D%7Bx_2%2B2%7D%3D-1

如此,证毕!

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