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我相信,这是最省力的算法了(2022新高考1卷圆锥曲线)

2022-07-06 10:01 作者:数学老顽童  | 我要投稿

(2022新高考Ⅰ,21)已知点A%5Cleft(%202%2C1%20%5Cright)%20在双曲线 C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E2%7D%7Ba%5E2-1%7D%3D1a%3E1)上,直线lCPQ两点,直线APAQ的斜率之和为0.

(1)求l的斜率;

(2)若%5Ctan%20%5Cangle%20PAQ%3D2%5Csqrt%7B2%7D,求%5Cbigtriangleup%20PAQ的面积.

解:(1)由题可知%5Cfrac%7B2%5E2%7D%7Ba%5E2%7D-%5Cfrac%7B1%5E2%7D%7Ba%5E2-1%7D%3D1

解得a%3D%5Csqrt%7B2%7D

所以C的方程为%5Cfrac%7Bx%5E2%7D%7B2%7D-y%5E2%3D1.

C的方程可改写为

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%2B4x-4%7D%7B2%7D-%5Cleft(%20y-1%20%5Cright)%20%5E2-2y%2B1%3D1

整理,得

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

l的方程为%5Ccolor%7Bred%7D%7Bm%5Cleft(%20x-2%20%5Cright)%20%2Bn%5Cleft(%20y-1%20%5Cright)%20%3D1%7D

C联立,得

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%7D%7B2%7D-%5Cleft(%20y-1%20%5Cright)%20%5E2%2B%5Cleft%5B%202%5Cleft(%20x-2%20%5Cright)%20-2%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,得

%5Cfrac%7B%5Cleft(%20x-2%20%5Cright)%20%5E2%7D%7B2%7D-%5Cleft(%20y-1%20%5Cright)%20%5E2%2B2m%5Cleft(%20x-2%20%5Cright)%20%5E2%2B%5Cleft(%202n-2m%20%5Cright)%20%5Cleft(%20x-2%20%5Cright)%20%5Cleft(%20y-1%20%5Cright)%20-2n%5Cleft(%20y-1%20%5Cright)%20%5E2%3D0

合并同类项

%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Cleft(%20x-2%20%5Cright)%20%5E2%2B%5Cleft(%202n-2m%20%5Cright)%20%5Cleft(%20x-2%20%5Cright)%20%5Cleft(%20y-1%20%5Cright)%20-%5Cleft(%202n%2B1%20%5Cright)%20%5Cleft(%20y-1%20%5Cright)%20%5E2%3D0

各项同时除以%5Cleft(%20x%20-2%5Cright)%5E2,得

%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%2B%5Cleft(%202n-2m%20%5Cright)%20%5Ccdot%20%5Cfrac%7By-1%7D%7Bx-2%7D-%5Cleft(%202n%2B1%20%5Cright)%20%5Cleft(%20%5Cfrac%7By-1%7D%7Bx-2%7D%20%5Cright)%20%5E2%3D0

最后调一下符号

%5Cleft(%202n%2B1%20%5Cright)%20%5Cleft(%20%5Cfrac%7By-1%7D%7Bx-2%7D%20%5Cright)%20%5E2%2B%5Cleft(%202m-2n%20%5Cright)%20%5Ccdot%20%5Cfrac%7By-1%7D%7Bx-2%7D-%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%3D0

因为

k_%7BAP%7D%2Bk_%7BAQ%7D%3D%5Cfrac%7By_1-1%7D%7Bx_1-2%7D%2B%5Cfrac%7By_2-1%7D%7Bx_2-2%7D%3D%5Cfrac%7B2n-2m%7D%7B2n%2B1%7D%3D0

所以%5Ccolor%7Bred%7D%7Bn%3Dm%7D

所以l的方程为

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

整理得,%5Ccolor%7Bred%7D%7Bx%2By-%5Cleft(%203%2B%5Cfrac%7B1%7D%7Bm%7D%20%5Cright)%3D0%7D.

易知l的斜率为%5Ccolor%7Bred%7D%7B-1%7D.

(2)设%5Cangle%20PAQ%3D2%5Ctheta%20%5Ccolor%7Bred%7D%7B%5Ctheta%20%5Cin%20%5Cleft(%200%2C%5Cfrac%7B%5Cmathrm%7B%5Cpi%7D%7D%7B4%7D%20%5Cright)%7D,则

%5Ctan%20%5Cangle%20PAQ%3D%5Ctan%20%202%5Ctheta%20%3D%5Cfrac%7B2%5Ctan%20%20%5Ctheta%7D%7B1-%5Ctan%20%5E2%5Ctheta%7D%3D2%5Csqrt%7B2%7D

解得%5Ccolor%7Bred%7D%7B%5Ctan%20%20%5Ctheta%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%7D,或%5Ctan%20%20%5Ctheta%3D-%5Csqrt%7B2%7D%20

不妨设k_%7BAQ%7D%3C0%3Ck_%7BAP%7D

AP的倾斜角为%5Ctheta%20

k_%7BAP%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D

则直线%5Ccolor%7Bred%7D%7BAP%7D%5Ccolor%7Bred%7D%7BC%7D的渐近线平行只有唯一交点%5Ccolor%7Bred%7D%7BA%7D%5Ccolor%7Bred%7D%7BP%7D重合,不合题意.

%5Ccolor%7Bred%7D%7BAP%7D的倾斜角为%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20-%5Ctheta%7D

所以%5Ccolor%7Bred%7D%7Bk_%7BAP%7D%3D%5Csqrt%7B2%7D%7D%5Ccolor%7Bred%7D%7Bk_%7BAQ%7D%3D-%5Csqrt%7B2%7D%7D

是时候画个图了:

由(1)可知

%5Ccolor%7Bred%7D%7Bk_%7BAP%7D%5Ccdot%20k_%7BAP%7D%3D%5Cfrac%7B-%5Cleft(%202m%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%7D%7B2n%2B1%7D%3D-2%7D

所以m-2n%3D%5Cfrac%7B3%7D%7B4%7D

又因为n%3Dm

二者联立,解得m%3D-%5Cfrac%7B3%7D%7B4%7D

所以l的方程为%5Ccolor%7Bred%7D%7Bx%2By-%5Cfrac%7B5%7D%7B3%7D%3D0%7D.

l与直线x%3D2交于H,易知%5Ccolor%7Bred%7D%7B%5Cvert%20AH%5Cvert%20%3D%5Cfrac%7B4%7D%7B3%7D%20%7D.

联立直线AP与直线l,消去y

解得x_1%3D%5Cfrac%7B10-4%5Csqrt%7B2%7D%7D%7B3%7D

联立直线AQ与直线l,消去y

解得x_2%3D%5Cfrac%7B10%2B4%5Csqrt%7B2%7D%7D%7B3%7D,所以

%5Cleft%7C%20x_2-x_1%20%5Cright%7C%3D%5Cfrac%7B10%2B4%5Csqrt%7B2%7D%7D%7B3%7D-%5Cfrac%7B10-4%5Csqrt%7B2%7D%7D%7B3%7D%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B8%5Csqrt%7B2%7D%7D%7B3%7D%7D

所以

S_%7B%5Cbigtriangleup%20PAQ%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B4%7D%7B3%7D%5Ctimes%20%5Cfrac%7B8%5Csqrt%7B2%7D%7D%7B3%7D%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B16%5Csqrt%7B2%7D%7D%7B9%7D%7D.

我相信,这是最省力的算法了(2022新高考1卷圆锥曲线)的评论 (共 条)

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