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两条直线,一个方程(2022新高考2卷圆锥曲线)

2022-07-05 00:54 作者:数学老顽童  | 我要投稿

(2022新高考Ⅱ,21)设双曲线C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3E0b%3E0)的右焦点为F%5Cleft(%202%2C0%20%5Cright)%20,渐近线方程为y%3D%5Cpm%20%5Csqrt%7B3%7Dx.

(1)求C的方程;

(2)经过F的直线与C的渐近线交于AB两点,点P%5Cleft(%20x_1%2Cy_1%20%5Cright)%20Q%5Cleft(%20x_2%2Cy_2%20%5Cright)%20C上,且x_1%3Ex_2%3E0y_1%3E0,过P且斜率为-%5Csqrt%7B3%7D的直线与过Q且斜率为%5Csqrt%7B3%7D的直线交于点M,从下面三个条件①②③中选择两个条件,证明另一个条件成立:

MAB上;②PQ%5Cparallel%20AB;③%5Cleft%7C%20AM%20%5Cright%7C%3D%5Cleft%7C%20BM%20%5Cright%7C.


解:(1)由题可知c%3D2

a%5E2%2Bb%5E2%3D4

又因为%5Cfrac%7Bb%7D%7Ba%7D%3D%5Csqrt%7B3%7D

解得a%3D1b%3D%5Csqrt%7B3%7D

故双曲线C的方程为x%5E2-%5Cfrac%7By%5E2%7D%7B3%7D%3D1.

(2)先画个图:

双曲线C的渐近线方程可化为x%5E2-%5Cfrac%7By%5E2%7D%7B3%7D%3D0

A%5Cleft(%20x_3%2Cy_3%20%5Cright)%20B%5Cleft(%20x_4%2Cy_4%20%5Cright)%20AB的中点为N

所以

x_%7B3%7D%5E%7B2%7D-%5Cfrac%7By_%7B3%7D%5E%7B2%7D%7D%7B3%7D%3D0

x_%7B4%7D%5E%7B2%7D-%5Cfrac%7By_%7B4%7D%5E%7B2%7D%7D%7B3%7D%3D0

由点差法可知(此处从略)

%5Cfrac%7By_N%7D%7Bx_N%7D%5Ccdot%20k_%7BAB%7D%3D3

k_%7BAB%7D%3D%5Cfrac%7B3x_N%7D%7By_N%7D.

直线PM的方程为

y-y_M%3D-%5Csqrt%7B3%7D%5Cleft(%20x-x_M%20%5Cright)%20

直线QM的方程为

y-y_M%3D%5Csqrt%7B3%7D%5Cleft(%20x-x_M%20%5Cright)%20

故直线PM%5Ccup%20QM的方程可写为

%5Cleft(%20y-y_M%20%5Cright)%20%5E2%3D3%5Cleft(%20x-x_M%20%5Cright)%20%5E2

%5Cleft(%20x-x_M%20%5Cright)%20%5E2-%5Cfrac%7B%5Cleft(%20y-y_M%20%5Cright)%20%5E2%7D%7B3%7D%3D0

双曲线C的方程可改写为

%5Cleft(%20x-x_M%20%5Cright)%20%5E2%2B2x_M%5Ccdot%20x-x_%7BM%7D%5E%7B2%7D-%5Cfrac%7B%5Cleft(%20y-y_M%20%5Cright)%20%5E2%2B2y_M%5Ccdot%20y-y_%7BM%7D%5E%7B2%7D%7D%7B3%7D%3D1

整理,得

%5Cleft(%20x-x_M%20%5Cright)%20%5E2%2B2x_M%5Ccdot%20x-x_%7BM%7D%5E%7B2%7D-%5Cfrac%7B%5Cleft(%20y-y_M%20%5Cright)%20%5E2%7D%7B3%7D-%5Cfrac%7B2y_M%7D%7B3%7D%5Ccdot%20y%2B%5Cfrac%7By_%7BM%7D%5E%7B2%7D%7D%7B3%7D%3D1

PM%5Ccup%20QM的方程代入上式,可得

2x_M%5Ccdot%20x-x_%7BM%7D%5E%7B2%7D-%5Cfrac%7B2y_M%7D%7B3%7D%5Ccdot%20y%2B%5Cfrac%7By_%7BM%7D%5E%7B2%7D%7D%7B3%7D%3D1

整理,得

x_M%5Ccdot%20x-%5Cfrac%7By_M%7D%7B3%7D%5Ccdot%20y-%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x_%7BM%7D%5E%7B2%7D-%5Cfrac%7By_%7BM%7D%5E%7B2%7D%7D%7B3%7D%2B1%20%5Cright)%20%3D0……(%5Cstar%20

由于P%5Cleft(%20x_1%2Cy_1%20%5Cright)%20Q%5Cleft(%20x_2%2Cy_2%20%5Cright)%20皆满足(%5Cstar%20),

%5Cstar%20即为直线PQ之方程

易知k_%7BPQ%7D%3D%5Cfrac%7B3x_M%7D%7By_M%7D.

现在再看这三个条件:

MAB上;

若②成立,则%5Cfrac%7Bx_M%7D%7By_M%7D%3D%5Cfrac%7Bx_N%7D%7By_N%7D,可理解为M在直线ON上,也可理解为:M在线段AB的中线上;

而③等价于:M在线段AB的中垂线上.

显然,以其中任意两个作为条件,都可以推出第三个(三线共点).


两条直线,一个方程(2022新高考2卷圆锥曲线)的评论 (共 条)

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