SOLUTION: A point moves so that its distance from the point (2,-1) is equal to its distance from the x-axis. Find the equation of the locus. a. x^2 - 4x - 4y + 4 = 0 b. y^2 - 4x

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A point moves so that its distance from the point (2,-1) is equal to its distance from the x-axis. Find the equation of the locus. a. x^2 - 4x - 4y + 4 = 0 b. y^2 - 4x       Log On


   

Question 1189327: A point moves so that its distance from the point (2,-1) is equal to its distance from the x-axis. Find the equation of the
locus.
a. x^2 - 4x - 4y + 4 = 0
b. y^2 - 4x - 2y + 5 = 0
c. x^2 - 4x - 2y + 5 = 0
d. x^2 + 4x - 2y + 5 = 0
Answer by ikleyn(52799) About Me  (Show Source):
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A point moves so that its distance from the point (2,-1) is equal to its distance from the x-axis.
Find the equation of the locus.
a. x^2 - 4x - 4y + 4 = 0
b. y^2 - 4x - 2y + 5 = 0
c. x^2 - 4x - 2y + 5 = 0
d. x^2 + 4x - 2y + 5 = 0
~~~~~~~~~~~~~~ 

An equation of the locus is

    sqrt%28%28x-2%29%5E2%2B%28y%2B1%29%5E2%29 = |y|.    (notice the absolute value sign in the right side !)


Square both sides of the equation

    (x-2)^2 + (y+1)^2 = y^2


Simplify

    x^2 - 4x + 4 + y^2 + 2y + 1 = y^2

    x^2 - 4x + 2y + 5 = 0.


ANSWER.  The equation of the locus is  x^2 - 4x + 2y + 5 = 0.

         No one of the listed in the options list.

Solved.