Question

求直线y=kx+3与直线y=(-2/k)x-1的交点轨迹方程.

Answer 1

由kx+3=-2x/k-1得交点x坐标：x=-4/(k+2/k)

由y=kx+3得：k=(y-3)/x,

将k值代入交点x坐标得：x=-4/[(y-3)/x+2x/(y-3)]=-4x(y-3)/[(y-3)^2+2x^2]

化简得：(y-3)^2+2x^2+4(y-3)=0

即：(y-1)^2+2x^2=4

此轨迹为椭圆.