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撕下伪装,还是“斜率之积为定值”的套路(2019北京卷圆锥曲线)

2022-07-12 11:57 作者:数学老顽童  | 我要投稿

(2019北京,18)已知抛物线Cx%5E2%3D-2py经过点%5Cleft(%202%2C-1%20%5Cright)%20.

(1)求抛物线C的标准方程及其准线方程;

(2)设O为原点,过抛物线C的焦点作斜率不为0的直线l交抛物线C于两点MN,直线y%3D-1分别交直线OMON于点A和点B.求证:以AB为直径的圆经过y轴上的两个定点.

解:(1)由题可知2%5E2%3D-2p%5Ccdot%20%5Cleft(%20-1%20%5Cright)%20

解得p%3D2

所以C的方程为x%5E2%3D-4y

其准线方程为y%3D1.

(2)先画个图

M%5Cleft(%20x_1%2Cy_1%20%5Cright)%20N%5Cleft(%20x_2%2Cy_2%20%5Cright)%20A%5Cleft(%20x_3%2C-1%20%5Cright)%20B%5Cleft(%20x_4%2C-1%20%5Cright)%20

设以AB为直径的圆与y轴交于T%5Cleft(%200%2Ct%20%5Cright)%20,则必有%5Coverrightarrow%7BTA%7D%5Ccdot%20%5Coverrightarrow%7BTB%7D%3D0

x_3x_4%2B%5Cleft(%20t%2B1%20%5Cright)%20%5E2%3D0

欲证t为定值,只需证x_3x_4为定值.

因为A在直线OM上,所以-1%3Dk_%7BOM%7D%5Ccdot%20x_3

x_3%3D-%5Cfrac%7B1%7D%7Bk_%7BOM%7D%7D

同理可得x_4%3D-%5Cfrac%7B1%7D%7Bk_%7BON%7D%7D

%5Cfrac%7B1%7D%7Bk_%7BOM%7D%5Ccdot%20k_%7BON%7D%7D%2B%5Cleft(%20t%2B1%20%5Cright)%20%5E2%3D0

故只需证%5Cfrac%7B1%7D%7Bk_%7BOM%7D%5Ccdot%20k_%7BON%7D%7D为定值即可.

l的方程为kx-y%3D1

C联立,得x%5E2%3D-4y%5Cleft(%20kx-y%20%5Cright)%20

整理得x%5E2%2B4kxy-4y%5E2%3D0

各项同除以y%5E2,得

%5Cleft(%20%5Cfrac%7Bx%7D%7By%7D%20%5Cright)%20%5E2%2B4k%5Ccdot%20%5Cfrac%7Bx%7D%7By%7D-4%3D0

所以%5Cfrac%7B1%7D%7Bk_%7BOM%7D%7D%5Ccdot%20%5Cfrac%7B1%7D%7Bk_%7BON%7D%7D%3D%5Cfrac%7Bx_1%7D%7By_1%7D%5Ccdot%20%5Cfrac%7Bx_2%7D%7By_2%7D%3D-4

所以-4%2B%5Cleft(%20t%2B1%20%5Cright)%20%5E2%3D0

解得t%3D-3,或t%3D1

故以AB为直径的圆恒过%5Cleft(%200%2C-3%20%5Cright)%20%5Cleft(%200%2C1%5Cright)%20.

撕下伪装,还是“斜率之积为定值”的套路(2019北京卷圆锥曲线)的评论 (共 条)

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