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及时盘点变量间的比例关系(2020课标Ⅱ圆锥曲线)

2022-08-07 21:38 作者:数学老顽童  | 我要投稿

(2020课标Ⅱ,19)已知椭圆C_1%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1a%3Eb%3E0)的右焦点F与抛物线C_2的焦点重合,C_1的中心与C_2的顶点重合.过F且垂直于x轴的直线交C_1AB两点,交C_2CD两点,且%5Cleft%7C%20CD%20%5Cright%7C%3D%5Cfrac%7B4%7D%7B3%7D%5Cleft%7C%20AB%20%5Cright%7C.

(1)求C_1的离心率;

(2)设MC_1C_2的公共点.若%5Cleft%7C%20MF%20%5Cright%7C%3D5,求C_1C_2的标准方程.

解:(1)画个图

C_2的标准方程为y%5E2%3D2pxp%3E0

由题可知

b%5E2%2Bc%5E2%3Da%5E2……

c%3D%5Cfrac%7Bp%7D%7B2%7D……

x%3D%5Cfrac%7Bp%7D%7B2%7D代入y%5E2%3D2px

y%3D%5Cpm%20p

所以%5Cleft%7C%20CD%20%5Cright%7C%3D2p.

x%3Dc代入%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1

y%3D%5Cpm%20%5Cfrac%7Bb%5E2%7D%7Ba%7D

所以%5Cleft%7C%20AB%20%5Cright%7C%3D%5Cfrac%7B2b%5E2%7D%7Ba%7D.

所以2p%3D%5Cfrac%7B4%7D%7B3%7D%5Ccdot%20%5Cfrac%7B2b%5E2%7D%7Ba%7D……

联立①、②、③,消去bp,得

2c%5E2%2B3ac-2a%5E2%3D0

2%5Cleft(%20%5Cfrac%7Bc%7D%7Ba%7D%20%5Cright)%20%5E2%2B3%5Ccdot%20%5Cfrac%7Bc%7D%7Ba%7D-2%3D0

解得%5Cfrac%7Bc%7D%7Ba%7D%3D-2),或%5Cfrac%7Bc%7D%7Ba%7D%3D%5Cfrac%7B1%7D%7B2%7D

所以C_1的离心率为%5Cfrac%7B1%7D%7B2%7D.

(2)由(1)可知

%5Ccolor%7Bred%7D%7Ba%3Ab%3Ac%3Ap%3D2%3A%5Csqrt%7B3%7D%3A1%3A2%7D

及时盘点变量间的比例关系,有助于快速理清思路,做到“破一点而破全局

a%3Dpb%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7Dp

所以C_1的方程可化为%5Cfrac%7Bx%5E2%7D%7Bp%5E2%7D%2B%5Cfrac%7B4y%5E2%7D%7B3p%5E2%7D%3D1

C_2联立,得

3%5Cleft(%20%5Cfrac%7Bx%7D%7Bp%7D%20%5Cright)%20%5E2%2B8%5Ccdot%20%5Cfrac%7Bx%7D%7Bp%7D-3%3D0

解得%5Cfrac%7Bx%7D%7Bp%7D%3D-3),或%5Cfrac%7Bx%7D%7Bp%7D%3D%5Cfrac%7B1%7D%7B3%7D

x%3D%5Cfrac%7Bp%7D%7B3%7D

所以%5Cleft%7C%20MF%20%5Cright%7C%3D%5Cfrac%7Bp%7D%7B3%7D%2B%5Cfrac%7Bp%7D%7B2%7D%3D5

解得p%3D6

所以C_1C_2的方程分别为

%5Cfrac%7Bx%5E2%7D%7B36%7D%2B%5Cfrac%7By%5E2%7D%7B27%7D%3D1y%5E2%3D12x.

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