SOLUTION: find the equation of the set of all points.P(x,y) that is equidistant from y-axis and (4,0)

Algebra ->  Graphs -> SOLUTION: find the equation of the set of all points.P(x,y) that is equidistant from y-axis and (4,0)       Log On


   



Question 966935: find the equation of the set of all points.P(x,y) that is equidistant from y-axis and (4,0)

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Perpendicular distance from (x,y) to y-axis is,
D%5B1%5D=x
Distance from (x,y) to (4,0) is,
D%5B2%5D=sqrt%28%28x-4%29%5E2%2By%5E2%29
Set them equal to each other.
sqrt%28%28x-4%29%5E2%2By%5E2%29=x
%28x-4%29%5E2%2By%5E2=x%5E2
x%5E2-8x%2B16%2By%5E2=x%5E2=0
y%5E2=8x-16
The solution is a parabola with symmetric about the x-axis.
.
.
.
.
.
.
.
Let's choose a point to verify the solution.
Let x=6, then y%5E2=48-16=32
So y=sqrt%2832%29 and y=-sqrt%2832%29
The distance from (6,0) to (4,sqrt%2832%29%29+and+%28%7B%7B%7B4,-sqrt%2832%29) is
D%5E2=%286-4%29%5E2%2B%28sqrt%2832%29%29%5E2
D%5E2=4%2B32
D%5E2=36
D=6
which equals the distance from the y-axis so the solution checks out.
P%28x%2Cy%29=(x,sqrt%288x-16%29) and (x,-sqrt%288x-16%29)