Question

求过点（0,2）的直线被椭圆X2+Y2=2所截弦中点的轨迹方程

Answer 1

设弦两端点坐标为(x1,y1),(x2.y2),弦中点坐标为(x,y).弦所在直线斜率为k

x1²+y1²=2

x2²+y2²=2

相减得(x1+x2)(x1-x2)+(y1+y2)(y1-y2)=0

y1-y2/x1-x2=-(x1+x2)/(y1+y2)

k=y-2/x

而k=-(x1+x2)/(y1+y2)

x1+x2=2x,

y1+y2=2y

则k=-x/y

故有k=y-2/x=-x/y

化简得：x²+y²-2y=0