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椭圆的内准圆

2022-07-25 13:16 作者:数学老顽童  | 我要投稿


  • 已知椭圆C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3Eb%3E0)及圆Ox%5E2%2By%5E2%3D%5Cfrac%7Ba%5E2b%5E2%7D%7Ba%5E2%2Bb%5E2%7D.AB为椭圆上两点,若OA%5Cbot%20OB,则直线AB与圆O相切.

证明:设直线AB的方程为mx%2Bny%3D1

与椭圆联立,得:

%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D%5Cleft(%20mx%2Bny%20%5Cright)%20%5E2

展开:

%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3Dm%5E2x%5E2%2B2mnxy%2Bn%5E2y%5E2

并项:

%5Cleft(%20n%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20y%5E2%2B2mn%5Ccdot%20xy%2B%5Cleft(%20m%5E2-%5Cfrac%7B1%7D%7Ba%5E2%7D%20%5Cright)%20%5Ccdot%20x%5E2%3D0

各项同除以x%5E2

%5Cleft(%20n%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7By%7D%7Bx%7D%20%5Cright)%20%5E2%2B2mn%5Ccdot%20%5Cfrac%7By%7D%7Bx%7D%2Bm%5E2-%5Cfrac%7B1%7D%7Ba%5E2%7D%3D0

由韦达定理可知:

k_%7BOA%7D%5Ccdot%20k_%7BOB%7D%3D%5Cfrac%7Bm%5E2-%5Cfrac%7B1%7D%7Ba%5E2%7D%7D%7Bn%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%7D%3D-1

即:m%5E2%2Bn%5E2%3D%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D

所以O到直线AB的距离为:

%5Cbegin%7Baligned%7D%09%5Cleft%7C%20OH%20%5Cright%7C%26%3D%5Cfrac%7B%5Cleft%7C%20m%5Ccdot%200%2Bn%5Ccdot%200-1%20%5Cright%7C%7D%7B%5Csqrt%7Bm%5E2%2Bn%5E2%7D%7D%5C%5C%09%26%3D%5Cfrac%7B1%7D%7B%5Csqrt%7Bm%5E2%2Bn%5E2%7D%7D%5C%5C%09%26%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D%7D%7D%5C%5C%09%26%3D%5Cfrac%7Bab%7D%7B%5Csqrt%7Ba%5E2%2Bb%5E2%7D%7D%5C%5C%5Cend%7Baligned%7D

故直线AB与圆O相切,证毕.

  • 已知椭圆C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3Eb%3E0)及圆Ox%5E2%2By%5E2%3D%5Cfrac%7Ba%5E2b%5E2%7D%7Ba%5E2%2Bb%5E2%7D.AB为椭圆上两点,若直线AB与圆O相切,则OA%5Cbot%20OB.

证明:设直线AB的方程为mx%2Bny%3D1

因为直线AB与圆O相切,所以O到直线AB的距离

%5Cbegin%7Baligned%7D%09%5Cleft%7C%20OH%20%5Cright%7C%26%3D%5Cfrac%7B%5Cleft%7C%20m%5Ccdot%200%2Bn%5Ccdot%200-1%20%5Cright%7C%7D%7B%5Csqrt%7Bm%5E2%2Bn%5E2%7D%7D%5C%5C%09%26%3D%5Cfrac%7B1%7D%7B%5Csqrt%7Bm%5E2%2Bn%5E2%7D%7D%5C%5C%09%26%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D%7D%7D%5C%5C%09%26%3D%5Cfrac%7Bab%7D%7B%5Csqrt%7Ba%5E2%2Bb%5E2%7D%7D%5C%5C%5Cend%7Baligned%7D

即:m%5E2%2Bn%5E2%3D%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D


联立直线AB与椭圆,得:

%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D%5Cleft(%20mx%2Bny%20%5Cright)%20%5E2

展开:

%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3Dm%5E2x%5E2%2B2mnxy%2Bn%5E2y%5E2

并项:

%5Cleft(%20n%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20y%5E2%2B2mn%5Ccdot%20xy%2B%5Cleft(%20m%5E2-%5Cfrac%7B1%7D%7Ba%5E2%7D%20%5Cright)%20%5Ccdot%20x%5E2%3D0

各项同除以x%5E2

%5Cleft(%20n%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7By%7D%7Bx%7D%20%5Cright)%20%5E2%2B2mn%5Ccdot%20%5Cfrac%7By%7D%7Bx%7D%2Bm%5E2-%5Cfrac%7B1%7D%7Ba%5E2%7D%3D0

由韦达定理可知:

k_%7BOA%7D%5Ccdot%20k_%7BOB%7D%3D%5Cfrac%7Bm%5E2-%5Cfrac%7B1%7D%7Ba%5E2%7D%7D%7Bn%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7Bb%5E2%7D-n%5E2%7D%7Bn%5E2-%5Cfrac%7B1%7D%7Bb%5E2%7D%7D%20%3D-1

所以:OA%5Cbot%20OB,证毕.

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