Question

A（1,0）,B（1,π）|MA|*|MB|=1,动点M的轨迹方程

Answer 1

假设M(X,Y)

那么|MA|=根号[(X-1)^2+Y^2]

|MB|=根号[(X-1)^2+(Y-π)^2]

所以

|MA|*|MB|=根号[[(X-1)^2+Y^2][(X-1)^2+(Y-π)^2]=1

[[(X-1)^2+Y^2][(X-1)^2+(Y-π)^2]=1

就写到这里吧,我感觉下面不需要撑开了,否则不简洁了