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橢圓中的“圓冪定理”

2022-09-07 13:43 作者:數(shù)學(xué)老頑童  | 我要投稿


如圖:過點P(不在橢圓%5CGamma%20上)的直線ABCD分別與橢圓%5CGamma%20%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3Eb%3E0)交于A、BC、D,則:

%5Cbbox%5B%23def%2C5px%2Cborder%3A2px%20solid%20%23FF6A6A%5D%7B%5Cfrac%7B%5Cleft%7C%20PA%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PB%20%5Cright%7C%7D%7Br_%7BAB%7D%5E%7B2%7D%7D%3D%5Cfrac%7B%5Cleft%7C%20PC%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PD%20%5Cright%7C%7D%7Br_%7BCD%7D%5E%7B2%7D%7D%7D

注1:其中r_%7BAB%7Dr_%7BCD%7D分別為直線AB、CD方向半徑.

注2:直線的方向半徑指的是與直線平行共線的半徑.

證明:設(shè)點P的坐標為%5Cleft(%20x_0%2Cy_0%20%5Cright)%20

直線AB、CD的傾斜角分別為%5Calpha%20、%5Cbeta%20

直線AB、CD的方向半徑分別為%5Cleft%7C%20OM%20%5Cright%7C、%5Cleft%7C%20ON%20%5Cright%7C,如圖:

設(shè)直線OM的參數(shù)方程為%5Cbegin%7Bcases%7D%09x%3Dt%5Ccos%20%20%5Calpha%20%2C%5C%5C%09y%3Dt%5Csin%20%20%5Calpha%20%2C%5C%5C%5Cend%7Bcases%7D

t為參數(shù)),

與橢圓%5CGamma%20聯(lián)立解得t_%7BM%7D%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B%5Ccos%20%5E2%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Csin%20%5E2%5Calpha%7D%7Bb%5E2%7D%7D,所以

r_%7BAB%7D%5E%7B2%7D%3D%5Cleft%7C%20OM%20%5Cright%7C%5E2%3Dt_%7BM%7D%5E%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B%5Ccos%20%5E2%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Csin%20%5E2%5Calpha%7D%7Bb%5E2%7D%7D,

設(shè)直線AB的參數(shù)方程為%5Cbegin%7Bcases%7D%09x%3Dx_0%2Bt%5Ccos%20%20%5Calpha%20%2C%5C%5C%09y%3Dy_0%2Bt%5Csin%20%20%5Calpha%20%2C%5C%5C%5Cend%7Bcases%7D

t為參數(shù)),

與橢圓%5CGamma%20聯(lián)立得

%5Cleft(%20%5Cfrac%7B%5Ccos%20%5E2%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Csin%20%5E2%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20t%5E2%2B%5Cleft(%20%5Cfrac%7B2x_0%5Ccos%20%20%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B2y_0%5Csin%20%20%5Calpha%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20t%2B%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D-1%3D0,

所以

%5Cbegin%7Baligned%7D%0A%09%5Cleft%7C%20PA%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PB%20%5Cright%7C%26%3D%5Cleft%7C%20t_1%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20t_2%20%5Cright%7C%3D%5Cleft%7C%20t_1t_2%20%5Cright%7C%5C%5C%0A%09%26%3D%5Cfrac%7B%5Cleft%7C%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D-1%20%5Cright%7C%7D%7B%5Cfrac%7B%5Ccos%20%5E2%5Calpha%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Csin%20%5E2%5Calpha%7D%7Bb%5E2%7D%7D%5C%5C%0A%09%5Cend%7Baligned%7D

所以%5Cfrac%7B%5Cleft%7C%20PA%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PB%20%5Cright%7C%7D%7Br_%7BAB%7D%5E%7B2%7D%7D%3D%5Cleft%7C%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D-1%20%5Cright%7C

同理可得%5Cfrac%7B%5Cleft%7C%20PC%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PD%20%5Cright%7C%7D%7Br_%7BCD%7D%5E%7B2%7D%7D%3D%5Cleft%7C%20%5Cfrac%7Bx_%7B0%7D%5E%7B2%7D%7D%7Ba%5E2%7D%2B%5Cfrac%7By_%7B0%7D%5E%7B2%7D%7D%7Bb%5E2%7D-1%20%5Cright%7C

所以%5Cfrac%7B%5Cleft%7C%20PA%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PB%20%5Cright%7C%7D%7Br_%7BAB%7D%5E%7B2%7D%7D%3D%5Cfrac%7B%5Cleft%7C%20PC%20%5Cright%7C%5Ccdot%20%5Cleft%7C%20PD%20%5Cright%7C%7D%7Br_%7BCD%7D%5E%7B2%7D%7D.

證畢.

橢圓中的“圓冪定理”的評論 (共 條)

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