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『圆锥曲线』来道参数方程练练计算力

2023-01-16 20:01 作者:げいしも_芸  | 我要投稿

题目:

如图,在椭圆%5CGamma%3A%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1上有一动点A,F_1,F_2为该椭圆的焦点,连接AF_1,AF_2分别交Γ与B、C点,取BC的中点D,求D运动轨迹的参数方程

求解:

A(x_0%2Cy_0)%EF%BC%8C%E5%85%B6%E4%B8%ADx_0%3Da%5Ccos%20t%2Cy_0%3Db%5Csin%20t%2Ct%5Cin%5B0%2C2%5Cpi%5D

不难得到:

AF_1%3D(x_0-c)y%3Dy_0(x-c)

为了解出B的坐标(x_1,y_1),将AF_1的解析式代回Γ的方程中,得到:

%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D%0A%5Cfrac%7Bx%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D%2B%5Cfrac%7B%5Cleft%5B%5Cfrac%7By_%7B0%7D%7D%7Bx_%7B0%7D-c%7D(x-c)%5Cright%5D%5E%7B2%7D%7D%7Bb%5E%7B2%7D%7D%3D1%20%5C%5C%0A%5Cfrac%7B%5Cleft%5B%5Cfrac%7Bx_%7B0%7D-c%7D%7By_%7B0%7D%7D%20y%2Bc%5Cright%5D%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D%2B%5Cfrac%7By%5E%7B2%7D%7D%7Bb%5E%7B2%7D%7D%3D1%0A%5Cend%7Barray%7D%5Cright.

整理得到:

%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D%0A%5Cfrac%7Bb%5E%7B2%7D%5Cleft(x_%7B0%7D-c%5Cright)%5E%7B2%7D%2Ba%5E%7B2%7D%20y_%7B0%7D%5E%7B2%7D%7D%7B%5Cleft(x_%7B0%7D-c%5Cright)%5E%7B2%7D%7D%20x%5E%7B2%7D-%5Cfrac%7B2%20a%5E%7B2%7D%20c%20y_%7B0%7D%5E%7B2%7D%7D%7B%5Cleft(x_%7B0%7D-c%5Cright)%5E%7B2%7D%7D%20x%2B%5Cfrac%7Ba%5E%7B2%7D%20c%5E%7B2%7D%20y_%7B0%7D%5E%7B2%7D%7D%7B%5Cleft(x_%7B0%7D-c%5Cright)%5E%7B2%7D%7D-a%5E%7B2%7D%20b%5E%7B2%7D%3D0%20%5C%5C%0A%5Cfrac%7Bb%5E%7B2%7D%5Cleft(x_%7B0%7D-c%5Cright)%5E%7B2%7D%2Ba%5E%7B2%7D%20y_%7B0%7D%5E%7B2%7D%7D%7By_%7B0%7D%5E%7B2%7D%7D%20y%5E%7B2%7D%2B%5Cfrac%7B2%20b%5E%7B2%7D%20c%5Cleft(x_%7B0%7D-c%5Cright)%7D%7By_%7B0%7D%7D%20y%2Bb%5E%7B2%7D%20c%5E%7B2%7D-a%5E%7B2%7D%20b%5E%7B2%7D%3D0%0A%5Cend%7Barray%7D%5Cright.

由韦达定理可得:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ax_0%2Bx_1%3D%5Cfrac%7Ba%5E2cy_0%5E2%7D%7Bb%5E2(x_0-c)%5E2%2Ba%5E2y_0%5E2%7D%0A%5Cnewline%0Ay_0%2By_1%3D-%5Cfrac%7Bb%5E2cy_0(x_0-c)%7D%7Bb%5E2(x_0-c)%5E2%2Ba%5E2y_0%5E2%7D%0A%5Cend%7Bmatrix%7D%5Cright.

于是得到:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ax_1%3D%5Cfrac%7Ba%5E2cy_0%5E2%7D%7Bb%5E2(x_0-c)%5E2%2Ba%5E2y_0%5E2%7D-x_0%0A%0A%5Cnewline%0Ay_1%3D-%5Cfrac%7Bb%5E2cy_0(x_0-c)%7D%7Bb%5E2(x_0-c)%5E2%2Ba%5E2y_0%5E2%7D-y_0%0A%5Cend%7Bmatrix%7D%5Cright.

同理可得点C的坐标(x_2,y_2):

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ax_2%3D-%5Cfrac%7Ba%5E2cy_0%5E2%7D%7Bb%5E2(x_0%2Bc)%5E2%2Ba%5E2y_0%5E2%7D-x_0%0A%0A%5Cnewline%0Ay_2%3D%5Cfrac%7Bb%5E2cy_0(x_0%2Bc)%7D%7Bb%5E2(x_0%2Bc)%5E2%2Ba%5E2y_0%5E2%7D-y_0%0A%5Cend%7Bmatrix%7D%5Cright.

于是,利用中点坐标公式我们很容易就能得到点D运动轨迹的参数方程:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ax(t)%3Da%5E2b%5E2c%5Csin%5E2t%5Cleft(%5Cfrac%7B1%7D%7Bb%5E2(a%5Ccos%20t-c)%5E2%2Ba%5E2b%5E2%5Csin%5E2t%7D-%5Cfrac%7B1%7D%7Bb%5E2(a%5Ccos%20t%2Bc)%5E2%2Ba%5E2b%5E2%5Csin%5E2t%7D%5Cright)-a%5Ccos%20t%0A%5Cnewline%0Ay(t)%3Db%5E3c%5Csin%20t%5Cleft(%5Cfrac%7Ba%5Ccos%20t%2Bc%7D%7Bb%5E2(a%5Ccos%20t%2Bc)%5E2%2Ba%5E2b%5E2%5Csin%5E2t%7D-%5Cfrac%7Ba%5Ccos%20t-c%7D%7Bb%5E2(a%5Ccos%20t-c)%5E2%2Ba%5E2b%5E2%5Csin%5E2t%7D%5Cright)-b%5Csin%20t%0A%5Cend%7Bmatrix%7D%5Cright.

化简得到:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%0Ax(t)%3D%5Cfrac%7B-ab%5E4%5Ccos%20t%7D%7B(a%5E2%2Bc%5E2)%5E2-4a%5E2c%5E2%5Ccos%5E2t%7D%0A%5Cnewline%0Ay(t)%3D%5Cfrac%7B-b%5E3(a%5E2%2Bc%5E2)%5Csin%20t%7D%7B4a%5E2c%5E2%5Csin%5E2t%2Bb%5E4%7D%0A%5Cend%7Bmatrix%7D%5Cright.

这便是点D运动轨迹的参数方程,其在坐标轴内的图像如下:

原题出处:BV1UV4y157aH

图表:https://www.desmos.com/calculator/vpxyeifwnr?lang=zh-CN

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