Question

数学求轨迹方程

已知直线系L1: ax-ky+k=0, L2: kx-y-1=0,a为常数,k为 参数 ,L1与L2交点的轨迹方程是??

Answer 1

L1: ax-ky+k=0, L2: kx-y-1=0, 消去参数k: 得: aX^2 = Y^2 -1 此即为L1与L2交点的轨迹方程

Answer 2

列方程组,求出交点出交点 x=..... y=..... 消去参数k既可

Answer 3

已知直线系L1: ax-ky+k=0, L2: kx-y-1=0,a为常数,k为参数 ,L1与L2交点的轨迹方程是?? 解：求交点：，由ax-ky+k=0, kx-y-1=0,得x=(2k)/(k^2-a),y=(k^2+a)/(k^2-a)