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定点问题的升级考法(2020新高考Ⅰ圆锥曲线)

2022-08-08 19:02 作者:数学老顽童  | 我要投稿

(2020新高考Ⅰ,22)已知椭圆C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3Eb%3E0)的离心率为%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D,且过点A%5Cleft(%202%2C1%20%5Cright)%20.

(1)求C的方程;

(2)点MNC上,且AM%5Cbot%20ANAD%5Cbot%20MND为垂足.证明:存在定点Q,使得%5Cleft%7C%20DQ%20%5Cright%7C为定值.

解:(1)由题可知%5Cbegin%7Bcases%7D%09%5Cfrac%7Bc%7D%7Ba%7D%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2C%5C%5C%09%5Cfrac%7B2%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%5E2%7D%7Bb%5E2%7D%3D1%2C%5C%5C%09b%5E2%2Bc%5E2%3Da%5E2%2C%5C%5C%5Cend%7Bcases%7D

解得a%3D%5Csqrt%7B6%7Db%3D%5Csqrt%7B3%7D

所以C的方程为%5Cfrac%7Bx%5E2%7D%7B6%7D%2B%5Cfrac%7By%5E2%7D%7B3%7D%3D1.

(2)先画个图

椭圆C的方程可化为

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%2B4x-4%7D%7B6%7D%2B%5Cfrac%7B%5Cleft(%20y-1%20%5Cright)%20%5E2%2B2y-1%7D%7B3%7D%3D1

整理,得

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%7D%7B6%7D%2B%5Cfrac%7B%5Cleft(%20y-1%20%5Cright)%20%5E2%7D%7B3%7D%2B%5Cfrac%7B2%7D%7B3%7D%5Cleft(%20x-2%20%5Cright)%20%2B%5Cfrac%7B2%7D%7B3%7D%5Cleft(%20y-1%20%5Cright)%20%3D0.

设直线MN的方程为

m%5Cleft(%20x-2%20%5Cright)%20%2Bn%5Cleft(%20y-1%20%5Cright)%20%3D1

与椭圆C联立(齐次化联立),得

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%7D%7B6%7D%2B%5Cfrac%7B%5Cleft(%20y-1%20%5Cright)%20%5E2%7D%7B3%7D%2B%5Cleft%5B%20%5Cfrac%7B2%7D%7B3%7D%5Cleft(%20x-2%20%5Cright)%20%2B%5Cfrac%7B2%7D%7B3%7D%5Cleft(%20y-1%20%5Cright)%20%5Cright%5D%20%5Cleft%5B%20m%5Cleft(%20x-2%20%5Cright)%20%2Bn%5Cleft(%20y-1%20%5Cright)%20%5Cright%5D%20%3D0

展开

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%7D%7B6%7D%2B%5Cfrac%7B%5Cleft(%20y-1%20%5Cright)%20%5E2%7D%7B3%7D%2B%5Cfrac%7B2m%7D%7B3%7D%5Cleft(%20x-2%20%5Cright)%20%5E2%2B%5Cfrac%7B2%5Cleft(%20m%2Bn%20%5Cright)%7D%7B3%7D%5Cleft(%20x-2%20%5Cright)%20%5Cleft(%20y-1%20%5Cright)%20%2B%5Cfrac%7B2n%7D%7B3%7D%5Cleft(%20y-1%20%5Cright)%20%5E2%3D0

并项

%5Cfrac%7B2n%2B1%7D%7B3%7D%5Ccdot%20%5Cleft(%20y-1%20%5Cright)%20%5E2%2B%5Cfrac%7B2%5Cleft(%20m%2Bn%20%5Cright)%7D%7B3%7D%5Cleft(%20x-2%20%5Cright)%20%5Cleft(%20y-1%20%5Cright)%20%2B%5Cfrac%7B4m%2B1%7D%7B6%7D%5Cleft(%20x-2%20%5Cright)%20%5E2%3D0

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

%5Cfrac%7B2n%2B1%7D%7B3%7D%5Ccdot%20%5Cleft(%20%5Cfrac%7By-1%7D%7Bx-2%7D%20%5Cright)%20%5E2%2B%5Cfrac%7B2%5Cleft(%20m%2Bn%20%5Cright)%7D%7B3%7D%5Ccdot%20%5Cfrac%7By-1%7D%7Bx-2%7D%2B%5Cfrac%7B4m%2B1%7D%7B6%7D%3D0

修饰一下,

%5Cleft(%204n%2B2%20%5Cright)%20%5Cleft(%20%5Cfrac%7By-1%7D%7Bx-2%7D%20%5Cright)%20%5E2%2B%5Cleft(%204m%2B4n%20%5Cright)%20%5Ccdot%20%5Cfrac%7By-1%7D%7Bx-2%7D%2B4m%2B1%3D0

所以

k_%7BAM%7D%5Ccdot%20k_%7BAN%7D%3D%5Cfrac%7B4m%2B1%7D%7B4n%2B2%7D%3D-1

所以n%3D-m-%5Cfrac%7B3%7D%7B4%7D

所以直线MN的方程为

m%5Cleft(%20x-2%20%5Cright)%20-%5Cleft(%20m%2B%5Cfrac%7B3%7D%7B4%7D%20%5Cright)%20%5Cleft(%20y-1%20%5Cright)%20%3D1

整理,得

m%5Cleft(%20x-y-1%20%5Cright)%20-%5Cfrac%7B3%7D%7B4%7D%5Cleft(%20y-1%20%5Cright)%20%3D1

x-y-1%3D0

-%5Cfrac%7B3%7D%7B4%7D%5Cleft(%20y-1%20%5Cright)%20%3D1

联立二者,解得%5Cbegin%7Bcases%7D%09x%3D%5Cfrac%7B2%7D%7B3%7D%5C%5C%09y%3D-%5Cfrac%7B1%7D%7B3%7D%5C%5C%5Cend%7Bcases%7D

所以直线MN过定点B%5Cleft(%20%5Cfrac%7B2%7D%7B3%7D%2C-%5Cfrac%7B1%7D%7B3%7D%20%5Cright)%20.

因为AD%5Cbot%20MN

所以AD%5Cbot%20BD

所以D在以AB为直径的圆上,

所以该圆心到D的距离为定值,

故该圆心即为所求之定点Q

QAB中点,可计算得

Q的坐标为%5Cleft(%20%5Cfrac%7B4%7D%7B3%7D%2C%5Cfrac%7B1%7D%7B3%7D%20%5Cright)%20.

定点问题的升级考法(2020新高考Ⅰ圆锥曲线)的评论 (共 条)

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