Locus in Mathematics: Exploring Definitions and Equations

The equation of the locus or loci is different from a regular geometrical equation as you need to derive the shape equation according to certain properties. The process by which you can derive the locus equation is:

• First, you have to assume the coordinates of the moving point of the locus as (X1, Y1)
• Then, you are to apply the geometrical condition on the two selected moving points, which helps you to obtain the relation between these two points. 
• In your result, you should replace X1 with x and Y1 with y. 
• The equation that you obtain after this will be your locus equation.

Examples

Common Loci Circles: x2+y2=2(x+y)

Find the equation of a point’s locus so that the sum of its distances from (0, -1) and (0,1) is 3. What exactly does the equation mean?

Solution

Suppose P(x, y) be a point on the given locus.

It is given that

A = (0, −1)

B = (0, 1)

From the conditions given,

Distance from Point P to A + Distance from Point P to B = 3

Hence, we can write it as

PA + PB = 3

By rearranging the above equation,

PA = 3 − PB

Now, Squaring both sides of the equation,

PA2=(3−PB)2

PA2 = 9 + PB2 − 6PB

x2+(y+1)2 = 9+x2+(y−1)2−6PB

(y+1)2−(y−1)2−9=−6PB

4y−9 = −6PB

Again, squaring both sides,

16y2−72y+81 = 36PB2

16y2−72y+81 = 36(x2+(y−1)2)

16y2−72y+81 = 36x2+36y2−72y+36

36x2+20y2 = 45

The above equation is written as follows

(x2/a2)+(y2/b2)=1

Hence, the above equation defines a circle

Therefore, the equation of locus is,

36x2+20y2 = 45, which is a circle.

Common Loci Lines: Y = MX + C

Find the locus of a point which is at a distance of 5 units from A(4,-3).

Solution:

Let P(x,y) be a point on the locus.

The given geometric condition is,

PA=5

√[(x-4)2 +(y+3)2] = 5

Squaring on both sides,

(x-4)2 +(y+3)2=25

x2 +16-8x+y2+9+6y=25

x2+y2-8x+6y=0 is the equation of locus.

Common Loci Parabola: y2+2ax+2by+c=0

Find the locus of a point which is equidistant from the points A(-3,2) and B(0,4).

Solution:

Let P(x,y) be a point on the locus.

The given geometric condition is,

PA=PB

√[(x+3)2 +(y-2)2] = √[(x-0)2 +(y-4)2]

Squaring on both sides,

(x+3)2 +(y-2)2 = (x-0)2 +(y-4)2

x2 +9+6x+y2+4-4y= x2 +y2+16-8y

Simplifying the above equation,

6x + 4y = 3, which is the required equation of locus.