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离心率只是比例,我们只要比例(2021浙江,16)

2023-04-23 10:46 作者:数学老顽童  | 我要投稿

(2021浙江,16)已知椭圆%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3Eb%3E0),焦点F_1%5Cleft(%20-c%2C0%20%5Cright)%20F_2%5Cleft(%20c%2C0%20%5Cright)%20%5Cleft(%20c%3E0%20%5Cright)%20.过F_1的直线和圆%5Cleft(%20x-%5Cfrac%7B1%7D%7B2%7Dc%20%5Cright)%20%5E2%2By%5E2%3Dc%5E2相切,与椭圆在第一象限交于点P,且PF_2%5Cbot%20x轴,则该直线的斜率是________,椭圆的离心率是________.

解:

设圆%5Cleft(%20x-%5Cfrac%7B1%7D%7B2%7Dc%20%5Cright)%20%5E2%2By%5E2%3Dc%5E2的圆心为M,与直线PF_1相切于点H.

易知%5Cvert%20MF_1%20%5Cvert%20%3Dc%2B%5Cfrac%7Bc%7D%7B2%7D%20%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B3c%7D%7B2%7D%20%7D.

又因为MH%5Cbot%20F_1H,且%5Cvert%20MH%5Cvert%20%3D%5Ccolor%7Bred%7Dc

所以%5Cleft%7C%20F_1H%20%5Cright%7C%3D%5Csqrt%7B%5Cleft(%20%5Cfrac%7B3%7D%7B2%7Dc%20%5Cright)%20%5E2-c%5E2%7D%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B2%7Dc%7D.

所以直线PF_1的斜率

%5Ccolor%7Bred%7D%7Bk_%7BPF_1%7D%3D%5Ctan%20%5Cmeasuredangle%20HF_1M%3D%5Cfrac%7B%5Cleft%7C%20HM%20%5Cright%7C%7D%7B%5Cleft%7C%20HF_1%20%5Cright%7C%7D%3D%5Cfrac%7Bc%7D%7B%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B2%7Dc%7D%3D%5Cfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D%7D

易知%5Cbigtriangleup%20F_1PF_2%5Ctext%7B%E2%88%BD%7D%5Cbigtriangleup%20F_1MH,故离心率

%5Cbegin%7Baligned%7D%0A%09e%26%3D%5Cfrac%7Bc%7D%7Ba%7D%3D%5Cfrac%7B2c%7D%7B2a%7D%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Cleft%7C%20F_1F_2%20%5Cright%7C%7D%7B%5Cleft%7C%20PF_1%20%5Cright%7C%2B%5Cleft%7C%20PF_2%20%5Cright%7C%7D%7D%5C%5C%0A%09%26%5Ccolor%7Bred%7D%7B%3D%5Cfrac%7B%5Cleft%7C%20F_1H%20%5Cright%7C%7D%7B%5Cleft%7C%20MF_1%20%5Cright%7C%2B%5Cleft%7C%20MH%20%5Cright%7C%7D%7D%3D%5Cfrac%7B%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B2%7Dc%7D%7B%5Cfrac%7B3%7D%7B2%7Dc%2Bc%7D%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Csqrt%7B5%7D%7D%7B5%7D%7D%5C%5C%0A%5Cend%7Baligned%7D

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