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点差复点差(2020浙江圆锥曲线)

2022-09-13 15:41 作者:数学老顽童  | 我要投稿

(2020浙江,21)如图,已知椭圆C_1%5Cfrac%7Bx%5E2%7D%7B2%7D%2By%5E2%3D1,抛物线C_2y%5E2%3D2pxp%3E0),点A是椭圆C_1与抛物线C_2的交点,过点A的直线l交椭圆C_1于点B,交抛物线C_2于点MBM不同于A

(1)若p%3D%5Cfrac%7B1%7D%7B16%7D,求抛物线C_2的焦点坐标;

(2)若存在不过原点的直线l使M为线段AB的中点,求p%0A的最大值.

解:(1)若p%3D%5Cfrac%7B1%7D%7B16%7D%20

%5Cfrac%7Bp%7D%7B2%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B16%7D%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B32%7D

C_2的焦点坐标为%5Cleft(%20%5Cfrac%7B1%7D%7B32%7D%2C0%20%5Cright)%20.

(2)分别设点ABM的坐标为%5Cleft(%20x_1%2Cy_1%20%5Cright)%20%5Cleft(%20x_2%2Cy_2%20%5Cright)%20%5Cleft(%20x_0%2Cy_0%20%5Cright)%20

因为ABC_1上,所以

%5Cfrac%7Bx_%7B1%7D%5E%7B2%7D%7D%7B2%7D%2By_%7B1%7D%5E%7B2%7D%3D1……①

%5Cfrac%7Bx_%7B2%7D%5E%7B2%7D%7D%7B2%7D%2By_%7B2%7D%5E%7B2%7D%3D1……②

-②(点差法),得

%5Cfrac%7Bx_%7B1%7D%5E%7B2%7D-x_%7B2%7D%5E%7B2%7D%7D%7B2%7D%2By_%7B1%7D%5E%7B2%7D-y_%7B2%7D%5E%7B2%7D%3D0

%5Cfrac%7By_1%2By_2%7D%7Bx_1%2Bx_2%7D%5Ccdot%20%5Cfrac%7By_1-y_2%7D%7Bx_1-x_2%7D%3D-%5Cfrac%7B1%7D%7B2%7D

%5Cfrac%7B2y_0%7D%7B2x_0%7D%5Ccdot%20%5Cfrac%7By_1-y_2%7D%7Bx_1-x_2%7D%3D-%5Cfrac%7B1%7D%7B2%7D

%5Ccolor%7Bred%7D%7B%5Cfrac%7By_0%7D%7Bx_0%7D%5Ccdot%20%7D%5Ccolor%7Bgreen%7D%7B%5Cfrac%7By_1-y_2%7D%7Bx_1-x_2%7D%7D%5Ccolor%7Bred%7D%7B%3D-%5Cfrac%7B1%7D%7B2%7D%7D……(%5Coplus

因为AMC_2上,所以

y_%7B1%7D%5E%7B2%7D%3D2px_1……③

y_%7B0%7D%5E%7B2%7D%3D2px_0……④

-④(点差法),得

y_%7B1%7D%5E%7B2%7D-y_%7B0%7D%5E%7B2%7D%3D2px_1-2px_0

%5Ccolor%7Bred%7D%7B%5Cleft(%20y_1%2By_0%20%5Cright)%20%5Ccdot%20%7D%5Ccolor%7Bgreen%7D%7B%5Cfrac%7By_1-y_0%7D%7Bx_1-x_0%7D%7D%5Ccolor%7Bred%7D%7B%3D2p%7D……(%5Cotimes%20

又因为%5Ccolor%7Bgreen%7D%7B%5Cfrac%7By_1-y_2%7D%7Bx_1-x_2%7D%7D%3D%5Ccolor%7Bgreen%7D%7B%5Cfrac%7By_1-y_0%7D%7Bx_1-x_0%7D%7D

所以%5Cfrac%7B%5Cotimes%20%7D%7B%5Coplus%20%7D%20%5Cleft(%20y_1%2By_0%20%5Cright)%20%5Ccdot%20%5Cfrac%7Bx_0%7D%7By_0%7D%3D-4p

将④代入上式,化简得

%5Ccolor%7Bred%7D%7By_%7B0%7D%5E%7B2%7D%2By_1%5Ccdot%20y_0%2B8p%5E2%3D0%7D.

所以%5CvarDelta%20%3Dy_%7B1%7D%5E%7B2%7D-4%5Ctimes%208p%5E2%5Cgeqslant%200

解得%5Ccolor%7Bred%7D%7By_%7B1%7D%5E%7B2%7D%5Cgeqslant%2032p%5E2%7D.

联立①、③,消去x_1%5Ccolor%7Bred%7D%7B%5Cfrac%7By_%7B1%7D%5E%7B4%7D%7D%7B8p%5E2%7D%2By_%7B1%7D%5E%7B2%7D%3D1%7D

所以%5Cfrac%7B%5Cleft(%2032p%5E2%20%5Cright)%20%5E2%7D%7B8p%5E2%7D%2B32p%5E2%5Cleqslant%201

解得p%5Cleqslant%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B40%7D

所以p的最大值为%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Csqrt%7B10%7D%7D%7B40%7D%7D.

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