Question

解析几何圆锥曲线弦中点的轨迹问题（求详细过程）

Answer 1

设BC的中点为(x,y),B(x+h,y+k),C(x-h,y-k),
A=π/2，A(0，4),
∴向量AB*AC=0，
∴(x+h)(x-h)+(y+k-4)(y-k-4)=0,
∴x^2-h^2+(y-4)^2-k^2=0,①
B,C在椭圆x^2/20+y^2/16=1上，
∴(x+h)^2/20+(y+k)^2/16=1,②
(x-h)^2/20+(y-k)^2/16=1.③
②-③，hx/5+ky/4=0,k=-4hx/(5y),④
代入②+③，(x^2+h^2)/10+[y^2+16h^2x^2/(25y^2)]/8=2,
∴(x^2+h^2)/10+(25y^4+16h^2x^2)/(200y^2)=2,
∴20x^2y^2+20h^2y^2+25y^4+16h^2x^2=400y^2,
∴(16x^2+20y^2)h^2=(400-20x^2-25y^2)y^2,
∴h^2=(400-20x^2-25y^2)y^2/(16x^2+20y^2),
代入④^2,k^2=16(400-20x^2-25y^2)x^2/[25(16x^2+20y^2)]
=4(80-4x^2-5y^2)x^2/[5(4x^2+5y^2)],
h^2+k^2=(80-4x^2-5y^2)(16x^2+25y^2)/[20(4x^2+5y^2)],
都代入①，x^2+(y-4)^2=(80-4x^2-5y^2)(16x^2+25y^2)/[20(4x^2+5y^2)],
去分母得20(4x^2+5y^2)(x^2+y^2-8y+16)=(80-4x^2-5y^2)(16x^2+25y^2),
整理得(4x^2+5y^2)(36x^2+45y^2-160y+320)=80(16x^2+25y^2),
9(4x^2+5y^2)^2-160y(4x^2+5y^2)-400y^2=0,
(4x^2+5y^2-20y)[9(4x^2+5y^2)+20y]=0,
∴4x^2+5y^2-20y=0，或9(4x^2+5y^2)+20y=0，
配方得x^2/5+(y-2)^2/4=1,或x^2/(5/81)+(y+2/9)^2/(4/81)=1,为所求.