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兩條直線,一個(gè)方程(2022新高考2卷圓錐曲線)

2022-07-05 00:54 作者:數(shù)學(xué)老頑童  | 我要投稿

(2022新高考Ⅱ,21)設(shè)雙曲線C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3E0,b%3E0)的右焦點(diǎn)為F%5Cleft(%202%2C0%20%5Cright)%20,漸近線方程為y%3D%5Cpm%20%5Csqrt%7B3%7Dx.

(1)求C的方程;

(2)經(jīng)過F的直線與C的漸近線交于A、B兩點(diǎn),點(diǎn)P%5Cleft(%20x_1%2Cy_1%20%5Cright)%20、Q%5Cleft(%20x_2%2Cy_2%20%5Cright)%20C上,且x_1%3Ex_2%3E0,y_1%3E0,過P且斜率為-%5Csqrt%7B3%7D的直線與過Q且斜率為%5Csqrt%7B3%7D的直線交于點(diǎn)M,從下面三個(gè)條件①②③中選擇兩個(gè)條件,證明另一個(gè)條件成立:

MAB上;②PQ%5Cparallel%20AB;③%5Cleft%7C%20AM%20%5Cright%7C%3D%5Cleft%7C%20BM%20%5Cright%7C.


解:(1)由題可知c%3D2,

a%5E2%2Bb%5E2%3D4

又因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Cfrac%7Bb%7D%7Ba%7D%3D%5Csqrt%7B3%7D" alt="%5Cfrac%7Bb%7D%7Ba%7D%3D%5Csqrt%7B3%7D">,

解得a%3D1b%3D%5Csqrt%7B3%7D,

故雙曲線C的方程為x%5E2-%5Cfrac%7By%5E2%7D%7B3%7D%3D1.

(2)先畫個(gè)圖:

雙曲線C的漸近線方程可化為x%5E2-%5Cfrac%7By%5E2%7D%7B3%7D%3D0

設(shè)A%5Cleft(%20x_3%2Cy_3%20%5Cright)%20、B%5Cleft(%20x_4%2Cy_4%20%5Cright)%20,AB的中點(diǎn)為N,

所以

x_%7B3%7D%5E%7B2%7D-%5Cfrac%7By_%7B3%7D%5E%7B2%7D%7D%7B3%7D%3D0

x_%7B4%7D%5E%7B2%7D-%5Cfrac%7By_%7B4%7D%5E%7B2%7D%7D%7B3%7D%3D0,

由點(diǎn)差法可知(此處從略)

%5Cfrac%7By_N%7D%7Bx_N%7D%5Ccdot%20k_%7BAB%7D%3D3

k_%7BAB%7D%3D%5Cfrac%7B3x_N%7D%7By_N%7D.

直線PM的方程為

y-y_M%3D-%5Csqrt%7B3%7D%5Cleft(%20x-x_M%20%5Cright)%20,

直線QM的方程為

y-y_M%3D%5Csqrt%7B3%7D%5Cleft(%20x-x_M%20%5Cright)%20,

故直線PM%5Ccup%20QM的方程可寫為

%5Cleft(%20y-y_M%20%5Cright)%20%5E2%3D3%5Cleft(%20x-x_M%20%5Cright)%20%5E2

%5Cleft(%20x-x_M%20%5Cright)%20%5E2-%5Cfrac%7B%5Cleft(%20y-y_M%20%5Cright)%20%5E2%7D%7B3%7D%3D0

雙曲線C的方程可改寫為

%5Cleft(%20x-x_M%20%5Cright)%20%5E2%2B2x_M%5Ccdot%20x-x_%7BM%7D%5E%7B2%7D-%5Cfrac%7B%5Cleft(%20y-y_M%20%5Cright)%20%5E2%2B2y_M%5Ccdot%20y-y_%7BM%7D%5E%7B2%7D%7D%7B3%7D%3D1,

整理,得

%5Cleft(%20x-x_M%20%5Cright)%20%5E2%2B2x_M%5Ccdot%20x-x_%7BM%7D%5E%7B2%7D-%5Cfrac%7B%5Cleft(%20y-y_M%20%5Cright)%20%5E2%7D%7B3%7D-%5Cfrac%7B2y_M%7D%7B3%7D%5Ccdot%20y%2B%5Cfrac%7By_%7BM%7D%5E%7B2%7D%7D%7B3%7D%3D1

PM%5Ccup%20QM的方程代入上式,可得

2x_M%5Ccdot%20x-x_%7BM%7D%5E%7B2%7D-%5Cfrac%7B2y_M%7D%7B3%7D%5Ccdot%20y%2B%5Cfrac%7By_%7BM%7D%5E%7B2%7D%7D%7B3%7D%3D1

整理,得

x_M%5Ccdot%20x-%5Cfrac%7By_M%7D%7B3%7D%5Ccdot%20y-%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x_%7BM%7D%5E%7B2%7D-%5Cfrac%7By_%7BM%7D%5E%7B2%7D%7D%7B3%7D%2B1%20%5Cright)%20%3D0……(%5Cstar%20

由于P%5Cleft(%20x_1%2Cy_1%20%5Cright)%20Q%5Cleft(%20x_2%2Cy_2%20%5Cright)%20皆滿足(%5Cstar%20),

%5Cstar%20即為直線PQ之方程,

易知k_%7BPQ%7D%3D%5Cfrac%7B3x_M%7D%7By_M%7D.

現(xiàn)在再看這三個(gè)條件:

MAB上;

若②成立,則%5Cfrac%7Bx_M%7D%7By_M%7D%3D%5Cfrac%7Bx_N%7D%7By_N%7D,可理解為M在直線ON上,也可理解為:M在線段AB的中線上;

而③等價(jià)于:M在線段AB的中垂線上.

顯然,以其中任意兩個(gè)作為條件,都可以推出第三個(gè)(三線共點(diǎn)).


兩條直線,一個(gè)方程(2022新高考2卷圓錐曲線)的評(píng)論 (共 條)

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