Question

求定点（c,0）（c>0)和它到定直线l:x=a^2/c距离之比是c/a(c/a>1)的点M的轨迹方程

Answer 1

设点M的坐标为(x,y)

|sqrt[(x-c)^2+y^2]|/|x-a^2/c|=c/a c>0,c/a>1

两边平方得

a^2[(x-c)^2+y^2]=c^2(x-a^2/c)^2

a^2x^2-2a^2cx+a^2c^2+a^2y^2=c^2x^2-2a^2cx+a^4

(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)

x^2/a^2-y^2/(c^2-a^2)=1

为双曲线.