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我相信,這是最省力的算法了(2022新高考1卷圓錐曲線(xiàn))

2022-07-06 10:01 作者:數(shù)學(xué)老頑童  | 我要投稿

(2022新高考Ⅰ,21)已知點(diǎn)A%5Cleft(%202%2C1%20%5Cright)%20在雙曲線(xiàn) C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E2%7D%7Ba%5E2-1%7D%3D1a%3E1)上,直線(xiàn)lCP、Q兩點(diǎn),直線(xiàn)AP、AQ的斜率之和為0.

(1)求l的斜率;

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

解:(1)由題可知%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的方程為%5Cfrac%7Bx%5E2%7D%7B2%7D-y%5E2%3D1.

C的方程可改寫(xiě)為

%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.

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

C聯(lián)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

展開(kāi),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

合并同類(lèi)項(xiàng)

%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,

各項(xiàng)同時(shí)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

最后調(diào)一下符號(hào)

%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?yàn)?/p>

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的方程為

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的斜率為%5Ccolor%7Bred%7D%7B-1%7D.

(2)設(shè)%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,則

%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

不妨設(shè)k_%7BAQ%7D%3C0%3Ck_%7BAP%7D,

AP的傾斜角為%5Ctheta%20

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

則直線(xiàn)%5Ccolor%7Bred%7D%7BAP%7D%5Ccolor%7Bred%7D%7BC%7D的漸近線(xiàn)平行,只有唯一交點(diǎn)%5Ccolor%7Bred%7D%7BA%7D、%5Ccolor%7Bred%7D%7BP%7D重合,不合題意.

%5Ccolor%7Bred%7D%7BAP%7D的傾斜角為%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,

是時(shí)候畫(huà)個(gè)圖了:

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?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=n%3Dm" alt="n%3Dm">,

二者聯(lián)立,解得m%3D-%5Cfrac%7B3%7D%7B4%7D,

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

設(shè)l與直線(xiàn)x%3D2交于H,易知%5Ccolor%7Bred%7D%7B%5Cvert%20AH%5Cvert%20%3D%5Cfrac%7B4%7D%7B3%7D%20%7D.

聯(lián)立直線(xiàn)AP與直線(xiàn)l,消去y,

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

聯(lián)立直線(xiàn)AQ與直線(xiàn)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,這是最省力的算法了(2022新高考1卷圓錐曲線(xiàn))的評(píng)論 (共 條)

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