Question

数学题求过点M（1,0）所作椭圆x^2/4+y^2=1的弦的中点的轨迹方程

Answer 1

解：简单用点差法＋共线求解。设弦交于椭圆于A(x1,y1),B(x2,y2),AB中点N(x,y)则x1+x2=2x,y1+y2=2y.A,B在椭圆上有x1^2/4+y1^2=1,x2^2/4+y2^2=1,当AB斜率k存在时两式相减得k=(y1-y2)/(x1-x2)=-(1/4)(x1+x2)/(y1+y2)=-(1/4)x/y,又A,B,M,N共线得k(AB)=k(MN)，有k=(y-0)/(x-1)=y/(x-1)=-(1/4)x/y,于是得中点N轨迹方程为y^2=-(1/4)x(x-1),(0=<x<=1)当AB斜率不存在时也满足此方程。仅参考。

Answer 2

解：设过点M的直线为x=my+1
代入方程x²+4y²=4
m²y²+2my+1+4y²=4
(m²+4)y²+2my-3=0
y1+y2=-2m/(m²+4)
x1+x2=m(y1+y2)+2=-2m²/(m²+4)+2=8/(m²+4)
设弦中点P（x，y）
x=(x1+x2)/2=4/(m²+4)
y=(y1+y2)/2=-m/(m²+4)
y/x=-m/4
x=my+1
m=(x-1)/y
所以
y/x=(1-x)/4y
x²-x+4y²=0
(x-1/2)²+4y²=1/4

Answer 3

y-0=k(x-1)
y=kx-k
代入x^2+2y^2=4
(1+2k^2)x^2-4k^2x+2k^2-4=0
x1+x2=4k^2/(1+2k^2)
y=kx-k
则y1+y2=k(x1+x2)-2k=-2k/(1+2k^2)

中点x=(x1+x2)/2,y=(y1+y2)/2
y/x=-2k/4k^2=-1/(2k)

y-0=k(x-1)
所以k=y/(x-1)
y/x=-(x-1)/2y
x^2-x+2y^2=0

Answer 4

一楼结果有误，第三步y^2系数应为是4.