Question 1191710: Derive the equation of the locus of a point P(x,y) which moves so that its distance from (2,3) is always equal to its distance from the line x+2=0
Found 3 solutions by josgarithmetic, ikleyn, greenestamps: Answer by josgarithmetic(39616) (Show Source): Answer by ikleyn(52776) (Show Source):
You can put this solution on YOUR website! .
Derive the equation of the locus of a point P(x,y) which moves so that
its distance from (2,3) is always equal to its distance from the line x+2=0.
~~~~~~~~~~~~~~~~
The line x+2 = 0 is the line x= -2 (vertical line parallel to y-axis with x-coordinate of -2).
Let (x,y) be the point of the locus. Then the distance from (x,y) to the line x= -2 is |x+2|
(notice the absolute value sign).
The distance from (x,y) to the point (2,3) is .
The equation of the locus is
= |x+2|.
Square both sides and get
(x-2)^2 + (y-3)^2 = (x+2)^2
x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 + 4x + 4
y^2 - 6y + 9 = 8x
(y-3)^2 = 8x
It is final equation of the locus. It represents a parabola with the horizontal axis y= 3,
parallel to x-axis. The parabola is opened right. Its vertex is the point (x,y) = (0,3).
Solved.
Answer by greenestamps(13198) (Show Source):
|
|
|