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一種相當(dāng)正經(jīng)(枯燥)的解法(2018北京圓錐曲線)

2022-07-18 13:59 作者:數(shù)學(xué)老頑童  | 我要投稿

(2018北京理,19)已知拋物線Cy%5E2%3D2px經(jīng)過點(diǎn)P%5Cleft(%201%2C2%20%5Cright)%20,過點(diǎn)Q%5Cleft(%200%2C1%20%5Cright)%20的直線l與拋物線C有兩個(gè)不同的交點(diǎn)A、B.且直線PAy軸于M,直線PBy軸于N.

(1)求直線l的斜率的取值范圍;

(2)設(shè)O為原點(diǎn),%5Coverrightarrow%7BQM%7D%3D%5Clambda%20%5Coverrightarrow%7BQO%7D,%5Coverrightarrow%7BQN%7D%3D%5Cmu%20%5Coverrightarrow%7BQO%7D,求證:%5Cfrac%7B1%7D%7B%5Clambda%7D%2B%5Cfrac%7B1%7D%7B%5Cmu%7D為定值.

解:(1)由2%5E2%3D2p%5Ccdot%201解得

p%3D2,

所以拋物線C的方程為y%5E2%3D4x

設(shè)l的方程為y%3Dkx%2B1,

與拋物線C聯(lián)立,得

k%5E2x%5E2%2B%5Cleft(%202k-4%20%5Cright)%20x%2B1%3D0,

顯然k%5Cne%200;

%5CvarDelta%20%3D%5Cleft(%202k-4%20%5Cright)%20%5E2-4k%5E2%5Ccdot%201%3D16%5Cleft(%201-k%20%5Cright)%20%3E0

所以k%3C1,

又因?yàn)橹本€PAPB皆與y軸相交,

所以直線PAPB必不過%5Cleft(%201%2C-2%20%5Cright)%20,

所以l亦不過%5Cleft(%201%2C2%20%5Cright)%20,

k%5Cne%20-3,

綜上所述:

k%5Cin%20%5Cleft(%20-%5Cinfty%20%2C-3%20%5Cright)%20%5Ccup%20%5Cleft(%20-3%2C0%20%5Cright)%20%5Ccup%20%5Cleft(%200%2C1%20%5Cright)%20

(2)先畫圖

由(1)知

x_1%2Bx_2%3D%5Cfrac%7B4-2k%7D%7Bk%5E2%7Dx_1x_2%3D%5Cfrac%7B1%7D%7Bk%5E2%7D,

直線PA的方程為

y-2%3D%5Cfrac%7By_1-2%7D%7Bx_1-1%7D%5Cleft(%20x-1%20%5Cright)%20,

y-2%3D%5Cfrac%7Bkx_1-1%7D%7Bx_1-1%7D%5Cleft(%20x-1%20%5Cright)%20,

x%3D0,可得

y_M%3D2-%5Cfrac%7Bkx_1-1%7D%7Bx_1-1%7D,

同理,y_N%3D2-%5Cfrac%7Bkx_2-1%7D%7Bx_2-1%7D

%5Coverrightarrow%7BQM%7D%3D%5Clambda%20%5Coverrightarrow%7BQO%7D知,

%5Cleft(%200%2Cy_M-1%20%5Cright)%20%3D%5Clambda%20%5Cleft(%200%2C-1%20%5Cright)%20

y_M-1%3D-%5Clambda,

%5Cfrac%7B1%7D%7B%5Clambda%20%20%7D%3D%5Cfrac%7B1%7D%7B1-y_M%7D,

同理,%5Cfrac%7B1%7D%7B%5Cmu%20%20%7D%3D%5Cfrac%7B1%7D%7B1-y_N%7D,

所以

%5Cbegin%7Baligned%7D%0A%09%5Cfrac%7B1%7D%7B%5Clambda%7D%2B%5Cfrac%7B1%7D%7B%5Cmu%7D%26%3D%5Cfrac%7B1%7D%7B1-%5Cleft(%202-%5Cfrac%7Bkx_1-1%7D%7Bx_1-1%7D%20%5Cright)%7D%2B%5Cfrac%7B1%7D%7B1-%5Cleft(%202-%5Cfrac%7Bkx_2-1%7D%7Bx_2-1%7D%20%5Cright)%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7Bk-1%7D%5Cleft(%202-%5Cfrac%7Bx_1%2Bx_2%7D%7Bx_1x_2%7D%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7Bk-1%7D%5Cleft(%202-%5Cfrac%7B%5Cfrac%7B4-2k%7D%7Bk%5E2%7D%7D%7B%5Cfrac%7B1%7D%7Bk%5E2%7D%7D%20%5Cright)%5C%5C%0A%09%26%3D2%5C%5C%0A%5Cend%7Baligned%7D



一種相當(dāng)正經(jīng)(枯燥)的解法(2018北京圓錐曲線)的評(píng)論 (共 條)

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