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面积分割术(2019课标Ⅲ圆锥曲线)

2022-08-23 20:33 作者:数学老顽童  | 我要投稿

(2019课标Ⅲ,21)已知曲线Cy%3D%5Cfrac%7Bx%5E2%7D%7B2%7DD为直线y%3D-%5Cfrac%7B1%7D%7B2%7D上的动点,过DC的两条切线,切点分别为AB.

(1)证明:直线AB过定点;

(2)若以E%5Cleft(%200%2C%5Cfrac%7B5%7D%7B2%7D%20%5Cright)%20为圆心的圆与直线AB相切,且切点为线段AB的中点,求四边形ADBE的面积.

解:(1)先画个图

设点ABD的坐标分别为

%5Cleft(%20x_1%2Cy_1%20%5Cright)%20B%5Cleft(%20x_2%2Cy_2%20%5Cright)%20D%5Cleft(%20k%2C-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20

求导,得y'%3Dx

故点A处的切线斜率为x_1

故点A处的切线方程为

y-y_1%3Dx_1%5Cleft(%20x-x_1%20%5Cright)%20

因该切线过点D

-%5Cfrac%7B1%7D%7B2%7D-y_1%3Dx_1%5Cleft(%20k-x_1%20%5Cright)%20

-%5Cfrac%7B1%7D%7B2%7D-y_1%3Dkx_1-x_%7B1%7D%5E%7B2%7D

-%5Cfrac%7B1%7D%7B2%7D-y_1%3Dkx_1-2y_1

%5Ccolor%7Bred%7D%7By_1%7D%3Dk%5Ccolor%7Bred%7D%7Bx_1%7D%2B%5Cfrac%7B1%7D%7B2%7D%20

同理可得%5Ccolor%7Bred%7D%7By_2%7D%3Dk%5Ccolor%7Bred%7D%7Bx_2%7D%2B%5Cfrac%7B1%7D%7B2%7D%20

可知点A%5Cleft(%20x_1%2Cy_1%20%5Cright)%20B%5Cleft(%20x_2%2Cy_2%20%5Cright)%20皆在直线

%5Ccolor%7Bred%7D%7By%7D%3Dk%5Ccolor%7Bred%7D%7Bx%7D%2B%5Cfrac%7B1%7D%7B2%7D%20上,

所以直线AB的方程即为

%5Ccolor%7Bred%7D%7By%7D%3Dk%5Ccolor%7Bred%7D%7Bx%7D%2B%5Cfrac%7B1%7D%7B2%7D%20.

易知其过定点F%5Cleft(%200%2C%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20.

(2)设线段AB的中点为M%5Cleft(%20x_0%2Cy_0%20%5Cright)%20

联立曲线C与直线AB,得

x%5E2-2kx-1%3D0

所以x_1%2Bx_2%3D2kx_1x_2%3D-1

所以x_0%3D%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%3Dk

所以y_0%3Dk%5E2%2B%5Cfrac%7B1%7D%7B2%7D

所以M的坐标为%5Cleft(%20k%2Ck%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20.

%5Cbigtriangleup%20EAB铅垂高%5Cleft%7C%20EF%20%5Cright%7C%3D2

%5Cbigtriangleup%20DAB铅垂高

%5Cleft%7C%20MD%20%5Cright%7C%3Dk%5E2%2B%5Cfrac%7B1%7D%7B2%7D-%5Cleft(%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%3Dk%5E2%2B1

而它们共同的水平宽

%5Cbegin%7Baligned%7D%0A%09%5Cleft%7C%20x_1-x_2%20%5Cright%7C%26%3D%5Csqrt%7B%5Cleft(%20x_1%2Bx_2%20%5Cright)%20%5E2-4x_1x_2%7D%5C%5C%0A%09%26%3D%5Csqrt%7B%5Cleft(%202k%20%5Cright)%20%5E2-4%5Ctimes%20%5Cleft(%20-1%20%5Cright)%7D%5C%5C%0A%09%26%3D2%5Csqrt%7Bk%5E2%2B1%7D%5C%5C%0A%5Cend%7Baligned%7D

所以

%5Cbegin%7Baligned%7D%0A%09S_%7B%5Ctext%7B%E5%9B%9B%E8%BE%B9%E5%BD%A2%7DADBE%7D%26%3DS_%7B%5Cbigtriangleup%20EAB%7D%2BS_%7B%5Cbigtriangleup%20DAB%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%7C%20x_1-x_2%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20EF%20%5Cright%7C%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%7C%20x_1-x_2%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20MD%20%5Cright%7C%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%7C%20x_1-x_2%20%5Cright%7C%5Ccdot%20%5Cleft(%20%5Cleft%7C%20EF%20%5Cright%7C%2B%5Cleft%7C%20MD%20%5Cright%7C%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%202%5Csqrt%7Bk%5E2%2B1%7D%5Ccdot%20%5Cleft(%202%2Bk%5E2%2B1%20%5Cright)%5C%5C%0A%09%26%3D%5Cleft(%20k%5E2%2B3%20%5Cright)%20%5Csqrt%7Bk%5E2%2B1%7D%5C%5C%0A%5Cend%7Baligned%7D

%5Coverrightarrow%7BFM%7D%3D%5Cleft(%20k%2Ck%5E2%20%5Cright)%20%5Coverrightarrow%7BEM%7D%3D%5Cleft(%20k%2Ck%5E2-2%20%5Cright)%20

由题可知%5Coverrightarrow%7BFM%7D%5Cbot%20%5Coverrightarrow%7BEM%7D

k%5Ccdot%20k%2Bk%5E2%5Ccdot%20%5Cleft(%20k%5E2-2%20%5Cright)%20%3D0

解得k%5E2%3D0,或k%5E2%3D1

k%5E2%3D0时,S_%7B%5Ctext%7B%E5%9B%9B%E8%BE%B9%E5%BD%A2%7DADBE%7D%3D3,如图:

k%5E2%3D1时,S_%7B%5Ctext%7B%E5%9B%9B%E8%BE%B9%E5%BD%A2%7DADBE%7D%3D4%5Csqrt%7B2%7D,如图:


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