Question 947097
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Let (x,y) be the point of the locus.


Then the distance to the point (0,2) is

    d = {{{sqrt((x-0)^2 + (y-2)^2)}}} = {{{sqrt(x^2 + (y-2)^2)}}}.


while the distance to the straight line y= 4 is  |y-4|.


Two distances are the same; it gives an equation

    {{{sqrt(x^2 + (y-2)^2)}}} = |y-4|.


Square both sides

    x^2 + y^2 - 4y + 4 = y^2 - 8y + 16.


Simplify

    x^2 = - 4y + 12,   or

    4y  = - x^2 + 12,  or

     y  = - {{{(1/4)x^2 + 3}}}.


It is the parabola opened downward, with the symmetry axis x= 0, 

with the vertex at the point  (0,3) and x-intercepts at x= - {{{2*sqrt(3)}}}  and  x= {{{2*sqrt(3)}}}.



    {{{graph(360,360,-6,6,-8,6,
       -(1/4)x^2 + 3, 4
)}}}


    Plot y = -{{{(1/4)x^2 + 3}}} (red) and y = 4 (green)
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Solved.