6.6 Equation of a Locus

6.6 Equation of a Locus

1. The equation of the locus of a moving point P (x, y) which is 
    always at a constant distance (r) from a fixed point (x1, y1) is:


2. The equation of the locus of a moving point P (x, y) which is 
     always at a constant distance from two fixed points (x1, y1) and 
     (x1, y1) with a ratio is:



3. The equation of the locus of a moving point P (x, y) which is 
    always equidistant from two fixed points A and B is the 
    perpendicular bisector of the straight line AB.


Example 1
Find the equation of the locus of a moving point P (x, y) which is always at a distance of 5 units from a fixed point Q (2, 4).

Solution:
(xx1)2 + (yy1)2 = r2
(x – 2)2 + (y – 4)2 = 52
x2 – 4x + 4 + y2 – 8y + 16 = 25
x2 + y2 – 4x – 8y – 5 = 0


Example 2
Find the equation of the locus of a moving point P (x, y) which is always equidistant from points A (-2, 3) and B (4, -1).

Solution:
Given PA=PB ( x( 2 ) ) 2 + ( y3 ) 2 = ( x4 ) 2 + ( y( 1 ) ) 2 Square both sides to eliminate the square roots. ( x+2 ) 2 + ( y3 ) 2 = ( x4 ) 2 + ( y+1 ) 2 x 2 +2x+4+ y 2 6y+9= x 2 8x+16+ y 2 +2y+1 10x8y4=0 Hence, the equation of the locus of point P is 10x8y4=0

Example 3
A (2, 0) and B (0, -2) are two fixed points. Point P moves with a ratio so that APPB = 1: 2.  Find the equation of the locus of point P.

Solution:
AP:PB=1:2 AP PB = 1 2 2AP=PB 2 ( x2 ) 2 + ( y0 ) 2 = ( x0 ) 2 + ( y( 2 ) ) 2 Square both sides to eliminate the square roots. 4[ ( x2 ) 2 + y 2 ]= x 2 + ( y+2 ) 2 4( x 2 4x+4+ y 2 )= x 2 + y 2 +4y+4 4 x 2 16x+16+4 y 2 = x 2 + y 2 +4y+4 3 x 2 +3 y 2 16x4y+12=0 Hence, the equation of the locus of point P is 3 x 2 +3 y 2 16x4y+12=0