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和抛物线有关,但关系不大(2019浙江圆锥曲线)

2022-09-08 22:21 作者:数学老顽童  | 我要投稿

(2019浙江,21)已知点F%5Cleft(%201%2C0%20%5Cright)%20为抛物线y%5E2%3D2pxp%3E0)的焦点.过点F的直线交抛物线于AB两点,点C在抛物线上,使得%5Cbigtriangleup%20ABC的重心Gx轴上,直线ACx轴于点Q,且点Q在点F的右侧,记%5Cbigtriangleup%20AFG%5Cbigtriangleup%20CQG的面积分别为S_1S_2.

(1)求p的值及抛物线的准线方程.

(2)求%5Cfrac%7BS_1%7D%7BS_2%7D的最小值及此时点G的坐标.

解:(1)由题可知%5Cfrac%7Bp%7D%7B2%7D%3D1

所以p%3D1

抛物线的准线方程为x%3D-1.

(2)设%5Coverrightarrow%7BAF%7D%3D%5Clambda%20%5Coverrightarrow%7BAB%7D%5Coverrightarrow%7BAQ%7D%3D%5Cmu%20%5Coverrightarrow%7BAC%7D

其中%5Clambda%20%5Cmu%20%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20

因为G%5Cbigtriangleup%20ABC的重心,所以

%5Cbegin%7Baligned%7D%0A%09%5Coverrightarrow%7BAG%7D%26%3D%5Cfrac%7B1%7D%7B3%7D%5Coverrightarrow%7BAB%7D%2B%5Cfrac%7B1%7D%7B3%7D%5Coverrightarrow%7BAC%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Clambda%7D%5Coverrightarrow%7BAF%7D%2B%5Cfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cfrac%7B1%7D%7B%5Cmu%7D%5Coverrightarrow%7BAQ%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B3%5Clambda%7D%5Coverrightarrow%7BAF%7D%2B%5Cfrac%7B1%7D%7B3%5Cmu%7D%5Coverrightarrow%7BAQ%7D%5C%5C%0A%5Cend%7Baligned%7D

又因为FGQ三点共线,所以

%5Ccolor%7Bred%7D%7B%5Cfrac%7B1%7D%7B3%5Clambda%7D%2B%5Cfrac%7B1%7D%7B3%5Cmu%7D%3D1%7D……(%5Cstar%20

因为%5Ccolor%7Bred%7D%7BS_%7B%5Cbigtriangleup%20ABG%7D%3DS_%7B%5Cbigtriangleup%20ACG%7D%7D,所以

%5Cbegin%7Baligned%7D%09%5Cfrac%7BS_1%7D%7BS_2%7D%26%3D%5Cfrac%7B%5Cfrac%7BS_1%7D%7BS_%7B%5Cbigtriangleup%20ABG%7D%7D%7D%7B%5Cfrac%7BS_2%7D%7BS_%7B%5Cbigtriangleup%20ACG%7D%7D%7D%3D%5Cfrac%7B%5Cfrac%7BAF%7D%7BAB%7D%7D%7B%5Cfrac%7BQC%7D%7BAC%7D%7D%3D%5Cfrac%7B%5Cfrac%7BAF%7D%7BAB%7D%7D%7B1-%5Cfrac%7BAQ%7D%7BAC%7D%7D%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Clambda%7D%7B1-%5Cmu%7D%7D%5C%5C%09%26%3D%5Cfrac%7B%5Cfrac%7B3%7D%7B1-%5Cfrac%7B1%7D%7B3%5Cmu%7D%7D%7D%7B1-%5Cmu%7D%3D%5Cfrac%7B1%7D%7B4-%5Cleft(%203%5Cmu%20%2B%5Cfrac%7B1%7D%7B%5Cmu%7D%20%5Cright)%7D%5C%5C%09%26%5Cgeqslant%20%5Cfrac%7B1%7D%7B4-2%5Csqrt%7B3%7D%7D%3D%5Ccolor%7Bred%7D%7B1%2B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%7D%5C%5C%5Cend%7Baligned%7D

当且仅当3%5Cmu%20%3D%5Cfrac%7B1%7D%7B%5Cmu%7D

%5Ccolor%7Bred%7D%7B%5Cmu%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D%7D时,取得最小值.

代入(%5Cstar%20),解得%5Ccolor%7Bred%7D%7B%5Clambda%20%3D%5Cfrac%7B1%7D%7B3-%5Csqrt%7B3%7D%7D%7D.

ABC三点的坐标分别为%5Cleft(%20x_1%2Cy_1%20%5Cright)%20%5Cleft(%20x_2%2Cy_2%20%5Cright)%20%5Cleft(%20x_3%2Cy_3%20%5Cright)%20,则

%5Ccolor%7Bred%7D%7B%5Cfrac%7By_1%7D%7By_1-y_2%7D%3D%5Cfrac%7B1%7D%7B3-%5Csqrt%7B3%7D%7D%7D……①,

注意,第二问到这一步为止都与抛物线无关


由(1)知抛物线的方程为y%5E2%3D4x

设直线AB的方程为x%3Dmy%2B1

两者联立得y%5E2-4my-4%3D0

所以%5Ccolor%7Bred%7D%7By_1y_2%3D-4%7D……②

联立①、②解得

%5Ccolor%7Bred%7D%7By_1%7D%3D%5Ccolor%7Bred%7D%7B%5Csqrt%7B2%7D%2B%5Csqrt%7B6%7D%7D%5Ccolor%7Bred%7D%7By_2%7D%3D%5Ccolor%7Bred%7D%7B%5Csqrt%7B2%7D-%5Csqrt%7B6%7D%7D,所以

%5Ccolor%7Bred%7D%7By_3%7D%3D-%5Cleft(%20y_1%2By_2%20%5Cright)%20%3D-%5Cleft(%20%5Csqrt%7B2%7D%2B%5Csqrt%7B6%7D%2B%5Csqrt%7B2%7D-%5Csqrt%7B6%7D%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B-2%5Csqrt%7B2%7D%7D

所以

%5Ccolor%7Bred%7D%7Bx_1%7D%3D%5Cfrac%7By_%7B1%7D%5E%7B2%7D%7D%7B4%7D%3D%5Cfrac%7B%5Cleft(%20%5Csqrt%7B2%7D%2B%5Csqrt%7B6%7D%20%5Cright)%20%5E2%7D%7B4%7D%3D%5Ccolor%7Bred%7D%7B2%2B%5Csqrt%7B3%7D%7D

%5Ccolor%7Bred%7D%7Bx_2%7D%3D%5Cfrac%7By_%7B2%7D%5E%7B2%7D%7D%7B4%7D%3D%5Cfrac%7B%5Cleft(%20%5Csqrt%7B2%7D-%5Csqrt%7B6%7D%20%5Cright)%20%5E2%7D%7B4%7D%3D%5Ccolor%7Bred%7D%7B2-%5Csqrt%7B3%7D%7D

%5Ccolor%7Bred%7D%7Bx_3%7D%3D%5Cfrac%7By_%7B3%7D%5E%7B2%7D%7D%7B4%7D%3D%5Cfrac%7B%5Cleft(%20-2%5Csqrt%7B2%7D%20%5Cright)%20%5E2%7D%7B4%7D%3D%5Ccolor%7Bred%7D%7B2%7D

所以

x_G%3D%5Cfrac%7Bx_1%2Bx_2%2Bx_3%7D%7B3%7D%3D%5Cfrac%7B2%2B%5Csqrt%7B3%7D%2B2-%5Csqrt%7B3%7D%2B2%7D%7B3%7D%3D2

所以G的坐标为%5Ccolor%7Bred%7D%7B%5Cleft(%202%2C0%20%5Cright)%20%7D.


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