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抛物线的平均性质(2022全国甲圆锥曲线)

2022-06-21 12:30 作者:数学老顽童  | 我要投稿

(2022全国甲,20)已知抛物线Cy%5E2%3D2pxp%3E0)的焦点为F,点D%5Cleft(%20p%2C0%20%5Cright)%20,过F的直线交CMN两点,当直线MD垂直于x轴时,%5Cvert%20MF%20%5Cvert%20%3D3.

(1)求C的方程;

(2)设直线MDNDC的另一个交点分别为AB,记直线MNAB的倾斜角分别为%5Calpha%20%5Cbeta%20.当%5Calpha-%5Cbeta%20取得最大值时,求直线AB的方程.

解:(1)当直线MD垂直于x轴时,

x%3Dp,可知y_D%3D%5Csqrt%7B2%7D%20p,所以%5Cvert%20MD%5Cvert%20%3D%5Csqrt%7B2%7D%20p

又因为%5Cleft%7C%20FD%5Cright%7C%3Dp-%5Cfrac%7Bp%7D%7B2%7D%3D%5Cfrac%7Bp%7D%7B2%7D

所以%5Cleft%7C%20MF%20%5Cright%7C%3D%5Csqrt%7B%5Cleft(%20%5Cfrac%7Bp%7D%7B2%7D%20%5Cright)%5E2%2B%5Cleft(%20%5Csqrt%7B2%7Dp%20%5Cright)%5E2%7D%3D3

解得p%3D2.

所以C的方程为y%5E2%3D4x.

(2)当直线AB的斜率不在时,易知直线MN的斜率也不存在,

此时%5Calpha%3D%5Cbeta%3D%5Cfrac%7B%5Cpi%7D%7B2%7D,所以%5Calpha-%5Cbeta%3D0

所以下面只需考虑%5Calpha-%5Cbeta%3E0的可能性.

如图,

A%5Cleft(%20x_1%2Cy_1%20%5Cright)B%5Cleft(%20x_2%2Cy_2%20%5Cright)

M%5Cleft(%20x_3%2Cy_3%20%5Cright)N%5Cleft(%20x_4%2Cy_4%20%5Cright)

设直线AB的方程为y%3Dk%5Cleft(%20x-t%20%5Cright)

C联立,得

k%5E2x%5E2-%5Cleft(%202tk%5E2%2B4%20%5Cright)%20x%2Bk%5E2t%5E2%3D0

所以x_1x_2%3Dt%5E2

所以%5Cfrac%7By_%7B1%7D%5E%7B2%7D%7D%7B4%7D%5Ccdot%20%5Cfrac%7By_%7B2%7D%5E%7B2%7D%7D%7B4%7D%3Dt%5E2

所以y_1y_2%3D-4t.

所以

x_1x_3%3D2%5E2%3D4

x_2x_4%3D2%5E2%3D4

y_1y_3%3D-4%5Ctimes%202%3D-8

y_2y_4%3D-4%5Ctimes%202%3D-8.

因为x_1x_2x_3x_4%3D16,且x_3x_4%3D1%5E2%3D1

所以x_1x_2%3D16,即t%5E2%3D16

解得t%3D4.

%5Cbegin%7Baligned%7D%20%09k_%7BMN%7D%26%3D%5Cfrac%7B-%5Cfrac%7B8%7D%7By_1%7D%2B%5Cfrac%7B8%7D%7By_2%7D%7D%7B%5Cfrac%7B4%7D%7Bx_1%7D-%5Cfrac%7B4%7D%7Bx_2%7D%7D%5C%5C%20%09%26%3D-2%5Ccdot%20%5Cfrac%7Bx_1x_2%7D%7By_1y_2%7D%5Ccdot%20%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%5C%5C%20%09%26%3D-2%5Ccdot%20%5Cfrac%7Bt%5E2%7D%7B-4t%7D%5Ccdot%20k%5C%5C%20%09%26%3D%5Cfrac%7Bt%7D%7B2%7D%5Ccdot%20k%5C%5C%20%09%26%3D%5Cfrac%7B4%7D%7B2%7D%5Ccdot%20k%5C%5C%20%09%26%3D2k%5C%5C%20%5Cend%7Baligned%7D

k%3C0,则k_%7BMN%7D%3D2k%3Ck%3C0

%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B2%7D%3C%5Calpha%20%3C%5Cbeta%20%3C%5Cmathrm%7B%5Cpi%7D

-%5Cfrac%7B%5Cpi%7D%7B2%7D%3C%5Calpha%20-%5Cbeta%20%3C0

k%3E0,则0%3Ck%3C2k%3Dk_%7BMN%7D

0%3C%5Cbeta%20%3C%5Calpha%20%3C%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B2%7D

0%3C%5Calpha%20-%5Cbeta%20%3C%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B2%7D

所以只考虑k%3E0的情形.

由于y%3D%5Ctan%20x%5Cleft(%200%2C%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)%5Cnearrow%20

所以当%5Ctan%20%5Cleft(%20%5Calpha%20-%5Cbeta%20%5Cright)最大时,%5Calpha-%5Cbeta最大.

%5Cbegin%7Baligned%7D%20%09%5Ctan%20%5Cleft(%20%5Calpha%20-%5Cbeta%20%5Cright)%20%26%3D%5Cfrac%7B%5Ctan%20%5Calpha%20-%5Ctan%20%5Cbeta%7D%7B1%2B%5Ctan%20%5Calpha%20%5Ctan%20%5Cbeta%7D%5C%5C%20%09%26%3D%5Cfrac%7Bk_%7BMN%7D-k%7D%7B1%2Bk_%7BMN%7D%5Ccdot%20k%7D%5C%5C%20%09%26%3D%5Cfrac%7B2k-k%7D%7B1%2B2k%5Ccdot%20k%7D%5C%5C%20%09%26%3D%5Cfrac%7Bk%7D%7B1%2B2k%5E2%7D%5C%5C%20%09%26%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B1%7D%7Bk%7D%2B2k%7D%5C%5C%20%26%5Cleqslant%20%5Cfrac%7B1%7D%7B2%5Csqrt%7B%5Cfrac%7B1%7D%7Bk%7D%5Ccdot%202k%7D%7D%5C%5C%20%26%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7B2%7D%7D%5C%5C%20%5Cend%7Baligned%7D

当且仅当%5Cfrac%7B1%7D%7Bk%7D%3D2k,即k%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D,取得最大值.

此时,直线AB的方程为y%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Cleft(%20x-4%20%5Cright)%20

整理得:x-%5Csqrt%7B2%7Dy-4%3D0.

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