Question

高二数学距离比它到直线距离小1轨迹方程问题

Answer 1

点P(x,y)到直线x+5=0的距离为,|x+5|,

点P到点E(4,0)的距离为,[(x-4)^2 + y^2]^(1/2),

[(x-4)^2 + y^2]^(1/2) = |x+5| -1 >=0,|x+5| >= 1,x=-4.

(x-4)^2 + y^2 = (x+5)^2 + 1 - 2|x+5|,

x^2 - 8x + 16 + y^2 = x^2 + 10x + 25 + 1 - 2|x+5|,

y^2 = 18x + 10 - 2|x+5|

x>=-4>-5时,y^2 = 18x + 10 - 2x - 10 = 16x,x>0.

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