Asked by austin
find the equation of the path of a point that moves so that its distance from (5, 0) is five~fourths of its distance from the line x=16\3
Answers
Answered by
Reiny
let the point be P(x,y)
If A is (5,0) and B = (16/3,y), then
AP = (5/4)BP
4 AP = 5 BP
4√( (x-5)^2 + (y-0)^2) = 5 (16/3 - x)
4√( (x-5)^2 + y^2 = 5(16/3-x)
square both sides
16( (x-5)^2 + y^2) = 25(16/3 - x)^2
16(x-5)^2 + 16y^2 = 25(16/3 - x)^2
looks hyperbolic to me.
check my algebra and arithmetic
If A is (5,0) and B = (16/3,y), then
AP = (5/4)BP
4 AP = 5 BP
4√( (x-5)^2 + (y-0)^2) = 5 (16/3 - x)
4√( (x-5)^2 + y^2 = 5(16/3-x)
square both sides
16( (x-5)^2 + y^2) = 25(16/3 - x)^2
16(x-5)^2 + 16y^2 = 25(16/3 - x)^2
looks hyperbolic to me.
check my algebra and arithmetic
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