SOLUTION: find the equation of the set of all points such that the differences of the distances of P(x,y) from (0,-3) and (0,3) is 1.

Question 62660: find the equation of the set of all points such that the differences of the distances of P(x,y) from (0,-3) and (0,3) is 1.
Answer by jai_kos(139) (Show Source): You can put this solution on YOUR website!
Given a point P(x, y), ( 0 , -3) and ( 0 ,3)
The difference of the distance between the point from P(x ,y) to ( 0 ,-3) and
the distance of the point from p(x ,y) to ( 0 ,3) is 1.
sqrt [(0 -x)^2 + (-3 - y)^2] - sqrt [(0-x) ^2 + ( 3-y)^2] = 1
==>Sqrt[ x^2 + 9 + y^2 + 6y ] = 1 + sqrt [ x^2 + 9 + y^2 -6y]
Squaring on both sides we get
{ Sqrt[ x^2 + 9 + y^2 + 6y ]}^2 = {1 + sqrt [ x^2 + 9 + y^2 -6y]} ^2
x^2 + 9 + y^2 + 6y = 1 + x^2 + 9 + y^2 -6y + 2 * 1 * sqrt [ x^2 + 9 + y^2 -6y]
6y + 6y - 1 = 2 sqrt(x^2 + 9 + y^2 -6y)
12y - 1 = 2 sqrt(x^2 + 9 + y^2 -6y)
Square on both sides, we get
(12y - 1)^2 = 4 (x^2 + 9 + y^2 -6y)
12y^2 + 1 -24y = 4x^2 + 36 + 4y^2 -24y
Bring all the terms to right hand side, we get
4x^2 -12y^2 + 4y^2 + 36 -1 = 0
4x^2 -8y^2 + 35 = 0
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