Locus of a variable point

Points with a geometrical property

The general method is illustrated by three examples.

Example 1

Take two fixed points A and B. Calculate the locus of the point P such that |PA| = 3.|PB|.

( |PA| means the distance from point P to point A. )


Since distance is a metric property, we choose orthonormal coordinate axes. We choose the midpoint of the segment [AB] as origin O and AB as x-axis. We choose the unity such that A(1,0); then we have B(-1,0). With this choice of the coordinate system, all calculations become easy. 

 
      P(x,y) is a point of the locus
<=>
       |PA| = 3.|PB|
<=>
      |PA|2 = 9.|PB|2
<=>
      (x-1)2 + y2 = 9 ((x + 1)2 +y2)
<=>
      ...
<=>
      2 x2 + 2 y2 + 5 x + 2 = 0

The last equation is the equation of a circle. It is called a circle of Appolonius. It is the locus of the point P such that |PA| = 3.|PB|.

Example 2

Take two points F and F' and a strictly positive value 2a such that 2a is greater than |F,F'|.
Calculate the locus of all points D such that |D,F| + |D,F'| = 2a


This locus is an ellipse.

Example 3

In a orthonormal coordinate system we take point A(a,0). The points P and P' are symmetric with respect to the y-axis. Calculate the locus of the variable point P such that 

 
   |PA|2 + |P'A|2 = 4 a2

 

 
      P(x,y) is a point of the locus
<=>
      |PA|2 + |P'A|2 = 4 a2
<=>
      (x - a)2 + y2 + (-x - a)2 + y2  = 4 a2
<=>
      ...
<=>
      x2 + y2 = a2

The last equation is the equation of a circle. It is the locus of the point P.

Points as intersection points of associated curves.

The general method is illustrated an example.

Example 4

The conic section with equation
m x2 + 2 xy + m y2 + 2m x - 2m y = 0
is variable because of the parameter m. Find the locus of the variable center-point.

 

 
    P(xo,yo) is center-point
<=>
    Fx' (xo,yo,zo) = 0 and Fy' (xo,yo,zo) = 0
<=>
    P(xo,yo) is intersection point of the lines
 
    Fx' (x,y,z) = 0 and  Fy' (x,y,z) = 0
 
<=>
    P(xo,yo) is intersection point of the lines
 
    m x + y + m = 0 and x + m y - m = 0

The equation of the locus appears if we eliminate the non homogeneous parameter m.
We find : 

 
    x2 - y2 + x + y = 0
 
<=>
    (x + y)(x - y + 1) = 0

The locus consists of two lines.

Singular parts of a locus

In some cases the associated lines can coincide for a particular value of the parameter.

Then, all points of these coinciding associated lines are points of the locus.

We say that these points form the singular part of the locus.

In previous example, the associated lines coincide for m = -1. The line (x - y + 1) = 0 is the singular part of the locus.

Parasitic part of a locus

We'll explain the concept by means of an example.

Example 5

In an orthonormal coordinate system, a circle has equation x2+y2=1 and A(2,0) is a fixed point.

A variable line through point A, intersects the circle in points B and C.

Calculate the locus of midpoint M of the segment [BC].


The variable line is y = m (x - 2).
The points B and C are the solutions of the system 

 
    x2 + y2 = 1
 
    y = m (x - 2)

The x-values are the roots x1 and x2 of 

 
(1 + m2) x2 - 4 m2 x + 4 m2 - 1 = 0

The y-values are the roots y1 and y2 of 

 
(1 + m2) y2 + 4 m y + 3 m2 = 0
 
The x-value of M is (x1 + x2)/2 = 2 m2/(1 + m2)
 
The y-value of M is (y1 + y2)/2 = - 2 m /(1 + m2)
 
So, we can say that M is the intersection point of the
associated lines
 
    x = 2 m2/(1 + m2)         (1)
 
    y = - 2 m /(1 + m2)        (2)

The locus arises from eliminating m from these equations. 

 
    (1)  <=>  (x - 2) m2 + x = 0
 
    (2)  <=>  y m2 + 2 m + y = 0

We can eliminate m with the help of the eliminant of sylvester. Then we find the equation of the locus: 

 
    x2 + y2 - 2x = 0
 
<=>
    (x - 1)2 + y2 = 1

The locus is a circle with midpoint (1,0) and radius = 1. But, when you draw several instances of the variable line through point A, there are many positions of that line that don't generate points B and C. From this we see that the locus is only that part of the circle that is inside or on the circle x2+y2=1. Therefore, we say that all the other points of the locus are parasitic points.

Many ways

In general, there are many ways to find a particular locus.

To illustrate this, we restart example 7 and we'll find the locus on a totaly different way. Here is the problem again.

In an orthonormal coordinate system, a circle has equation
x2 + y2 = 1 and A(2,0) is a fixed point.

A variable line through point A, intersects the circle in points B and C.

Calculate the locus of midpoint M of the segment [BC].


The variable line is y = m (x - 2).
The polar line of point P(1,m,0) with respect to the circle is x + my = 0 . This line cuts the variable line in point M.
The locus arises from eliminating m from these equations.
We find : x2 + y2 - 2x = 0.

Vectors and the parametric equation of a locus

Recall

P.Q = ||P||.||Q||. cos(t) 

·                ·          
·                ·           P = Q  <=> (x,y) = (x',y') <=> P.E1 = Q.E2

The cycloid

The cycloid is the trace of a point P on a circle rolling upon a line l without slipping.

Choose the X-axis on the line l.
Choose the origin of the orthonormal coordinate system in point P, on the moment P is on the line l.

For each position of the circle, we have (vectors in bold) 

 
    P = Q + QA + AP
<=>
    P.E1 = Q.E1 + QA.E1 + AP.E1
    P.E2 = Q.E2 + QA.E2 + AP.E2
<=>
    x = r.t + 0 + r.cos(-pi/2 - t)
    y =  0 +  r + r.cos(-pi - t)
<=>
    x = r(t - sin(t))
    y = r(1 - cos(t))

The last equations are the parametric equations of the cycloid.

The astroid

The astroid is the trace of a point P on a circle with radius R/4, rolling without slipping inside an other circle with radius R.

The small circle start rolling in point (R,0). Then t=s=0. Since the circle is rolling without slipping, we have that the two blue arcs have the same length.

 
   So, R.t = -R.s/4 <=>  s = - 4 t
 
   P = A + AP
=>
   P.E1 = A.E1 + AP.E1
<=>
    x = (3R/4).1.cos(t) + (R/4).1.cos(t+s)
<=>
    x = (3R/4).cos(t) + (R/4).cos(3t)             (1)
 
and in the same way
 
   P.E2 = A.E2 + AP.E2
=>
   y =  (3R/4).1.cos(pi/2 -t) + (R/4).1.cos(t - pi/2 + s)
<=>
   y =  (3R/4).sin(t) + (R/4).sin(3t)             (2)
 
Since cos(3t) = 4 cos3(t) - 3cos(t)
      sin(3t) = 3 sin(t) - 4 sin3(t)
 
we find the parametric equations of the astroid from (1) and (2)
 
    x = R cos3(t)
    y = R sin3(t)

The cartesian equation can be found through elimination of t. 

 
    cos(t) = (x/R)1/3
    sin(t) = (y/R)1/3
 
=>  cos2(t) + sin2(t)  = (x/R)2/3 +  (y/R)2/3
 
<=>  x2/3 + y2/3 = R2/3