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参数方程,新高考还能用吗?(2021新高考Ⅰ圆锥曲线)

2022-08-07 14:59 作者:数学老顽童  | 我要投稿

(2021新高考Ⅰ,21)在平面直角坐标系xOy中,已知点F_1%5Cleft(%20-%5Csqrt%7B17%7D%2C0%20%5Cright)%20F_2%5Cleft(%20%5Csqrt%7B17%7D%2C0%20%5Cright)%20,点M满足%5Cleft%7C%20MF_1%20%5Cright%7C-%5Cleft%7C%20MF_2%20%5Cright%7C%3D2.记M的轨迹为C.

(1)求C的方程;

(2)设点T在直线x%3D%5Cfrac%7B1%7D%7B2%7D%20上,过T的两条直线分别交CAB两点和PQ两点,且%5Cleft%7C%20TA%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20TB%20%5Cright%7C%3D%5Cleft%7C%20TP%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20TQ%20%5Cright%7C,求直线AB的斜率与直线PQ的斜率之和.


解:(1)易知C是以F_1F_2为焦点的双曲线的右支

2a%3D2

所以a%3D1

又因c%3D%5Csqrt%7B17%7D

所以b%5E2%3Dc%5E2-a%5E2%3D17-1%3D16

所以C的方程为x%5E2-%5Cfrac%7By%5E2%7D%7B16%7D%3D1x%5Cgeqslant%201

(2)先画个图

T的坐标为%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%2Ct%20%5Cright)%20

设直线ABPQ的倾斜角分别为%5Calpha%20%5Cbeta%20

设直线AB的参数方程为

%5Cbegin%7Bcases%7D%09x%3D%5Cfrac%7B1%7D%7B2%7D%2B%5Clambda%20%5Ccos%20%20%5Calpha%20%2C%5C%5C%09y%3Dt%2B%5Clambda%20%5Csin%20%20%5Calpha%20%2C%5C%5C%5Cend%7Bcases%7D%5Clambda%20为参数)

C的方程联立,得

%5Cleft(%2017%5Ccos%20%5E2%5Calpha%20-1%20%5Cright)%20%5Clambda%20%5E2%2B%5Cleft(%2016%5Ccos%20%20%5Calpha%20-2t%5Csin%20%20%5Calpha%20%5Cright)%20%5Clambda%20-t%5E2-12%3D0

所以

%5Cleft%7C%20TA%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20TB%20%5Cright%7C%3D%5Clambda%20_1%5Clambda%20_2%3D%5Cfrac%7Bt%5E2%2B12%7D%7B1-17%5Ccos%20%5E2%5Calpha%7D

同理可得

%5Cleft%7C%20TP%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20TQ%20%5Cright%7C%3D%5Cfrac%7Bt%5E2%2B12%7D%7B1-17%5Ccos%20%5E2%5Cbeta%7D.

所以

%5Cfrac%7Bt%5E2%2B12%7D%7B1-17%5Ccos%20%5E2%5Calpha%7D%3D%5Cfrac%7Bt%5E2%2B12%7D%7B1-17%5Ccos%20%5E2%5Cbeta%7D

化简得%5Ccos%20%5E2%5Calpha%20%3D%5Ccos%20%5E2%5Cbeta%20

所以%5Ccos%20%20%5Cbeta%20%3D%5Ccos%20%20%5Calpha%20

%5Ccos%20%20%5Cbeta%20%3D-%5Ccos%20%20%5Calpha%20

所以%5Cbeta%20%3D%5Cmathrm%7B%5Cpi%7D-%5Calpha%20

所以%5Ctan%20%20%5Cbeta%20%3D%5Ctan%20%5Cleft(%20%5Cmathrm%7B%5Cpi%7D-%5Calpha%20%5Cright)%20%3D-%5Ctan%20%20%5Calpha%20

k_%7BAB%7D%3D-k_%7BPQ%7D

所以k_%7BAB%7D%2Bk_%7BPQ%7D%3D0.

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