Loci




Definition

A locus is the set of all points that satisfy a given condition or a set of given conditions.

All examples here deal about points in a plane.

In the examples below we work with an orthonormal coordinate system. The choice of this coordinate system is very important because the difficulty of the calculations depends strongly on this choice.

First kind -- Starting from a geometric property

Example m1

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


We choose the coordinate system so that A(1,0) and B(-1,0).

 
      P(x,y) is a point on 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 an equation of a circle. Each point of the circle has the requested property, and vice versa.

Example m2

Find the locus of the points equally distant from two intersecting lines a and b.


We choose the coordinate system so that a and b are symmetric relative to the x-axis. The lines a and b have equations y = m x and y = - m x.
 
      P(x,y) is a point on the locus
<=>
       |P,a| = |P,b|
<=>
      |P,a|2 = |P,b|2
<=>
     (y - m x)2      (y + m x)2
     ------------ = --------------
       1 + m2          1 + m2
<=>

      (y - m x)2   =   (y + m x)2

<=>
             x.y = 0
<=>
        x= 0 or y = 0
We find, as expected, x-axis and y-axis as locus

Example m3

Point A is a fixed point A(a,0). P is a variable point in the plane. P 'is the mirror image of P relative to the y-axis. (draw a figure)

Find the locus of point P such that |PA|2 + |P'A|2 = 4a2.


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

Example m4

Consider the hyperbola y = 1/x. Through a variable point P (x, y) of the plane, we draw a line parallel to the x-axis and another line parallel to the y-axis. These lines intersect the hyperbola respectively in A and B. Find the locus of point P such that the centroid of triangle PAB is on the line x= 1.

The coordinates of P are (x,y)
The coordinates of A are (x, 1/x)
The coordinates of B are (1/y, y)

The x-value of the centroid is ( x + x + 1/y )/3 = (2x+ 1/y)/3
P(x,y) is a point on the locus if and only if (2x+ 1/y)/3 = 1

The locus has equation y = 1/(-2x+3).
It is the equation of a homographic function. It is a hyperbola with asymptotes y=0 and x=3/2. You can find more info abaut the homographic function via this link .

Analytical versus Geometric

A triangle ABC has a fixed side [AB] with length c.
Find the locus of the third vertex C such that the medians from A en C are orthogonal.
Next, find the maximum area of such a triangle.

  1. Analytical

    We choose an orthonormal coordinate system such that A(-c/2,0), B(c/2,0).
    C(x,y) is the variable third vertex.
    Make a figure.
    The center of [BC] is M( (2x+c)/4, y/2 ).
    The median from C has a slope y/x.
    The median AM has a slope 2y/(2x+3c).

     
        C(x,y) is a point of the locus
    <=>
        The medians from A and C are orthogonal
    <=>
           y        2y
          ---. ----------- = -1
           x     2x + 3c
    <=>
          2 y2 + 2x2 + 3c x = 0
    <=>
    
          x2 + y2 + (3c/2) x = 0
    <=>
          (x +  3c/4)2 + y2 = 9c2/16
    
         The locus of the vertex C is a circle
         with center (-3c/4,0) and radius 3c/4
    
    The area is maximum if and only if the height of the triangle is maximum. The maximum height is 3c/4. The maximum area is 3c2/8.
     
            
    
  2. Geometric:

    Say Z is the centroid of the variable triangle ABC. A and B are fixed points.
    The distance|AB| = c. Point O is the center of [AB].

     
       C(x,y) is a point of the locus
    <=>
        The medians AZ and CZ are orthogonal
    <=>
        The triangle AOZ is a right-angled triangle
    
    The fixed segment [AO] is the hypotenuse of the variable right-angled triangle AOZ.
    So, the locus of Z is the circle C1 with center (-c/4,0). The radius is c/2.

    But Z is the centroid of the triangle. So, the vector C = 3Z.
    From this, it follows that the locus of point C is a circle C2. It is the image of the transformation of circle C1 by a homothety with center O and factor 3.
    The center of the circle C2 is (-3c/4,0) and the radius is 3c/2.

Calculation of loci by associated curves

The method is shown using several examples.

It is assumed that you are aware of the elimination methods. These methods are explained here.

Example a1

Take a parabola P and a line b with equations
 
        /  y = x2  + mx
        \  y = 2x + m
m is a variable parameter.
For each fixed value of m, the plots P and b are fixed. Therefore, we say that P and b are associated curves. If m changes smoothly, the curves P and b change smoothly. The intersection points of P and b will also change smoothly. The intersection points describe a curve c. c is the locus.
 
   P(x,y) is a point of c

<=> P(x,y) is an intersection point of P and b

<=>  There is a value of m such that

        /  y = x2  + mx
        \  y = 2x + m

<=>  The following system has a solution for m

        / mx = y - x2
        \ m  = y - 2x
                        (elimination theory)
<=>
        | 1     y-2x |
        |            | = 0
        | x   y - x2 |

<=>     x2  - xy + y = 0

If we omit the intermediate calculations here, we see better what we found:
 
   P(x,y) is a point of c

<=>     x2  - xy + y = 0

The last equation is the equation of the locus c.

The equation of the locus arises by eliminating the parameter from the system formed by the associated curves.

Example a2

Take the lines a and b with equations
 
        / l x - m y = 0
        \ m x + l y = 5l
The real numbers l and m are parameters, not both 0. If we replace l and m with r.l and r.m the lines does not change.
Because of this property, we say that l and m are homogeneous parameters. The lines a and b are associated lines.
The intersection point describes the locus c.
 

        P(x,y) is on  c

<=>     P(x,y) is an intersection point of  a and  b

<=>     There is a value of  l and  m ( not both  0) such that
                l x - m y = 0
                m x + l y = 5l

<=>     The following system has a solution for l and m ( not both 0)

                x l - y m = 0
          (y - 5) l + x m = 0
                                       (elimination theory)

<=>             |  x     -y |
                |           |  = 0
                |y - 5    x |

<=>             x2  + y2  - 5 y = 0

The last equation is the equation of the locus c. The locus is a circle. The associated lines intersect on this circle. Each point of the circle is an intersection point of the two variable lines.

The equation of the locus arises by eliminating the parameters from the system formed by the associated lines.

Example a3

Take a circle with radius 2. Consider a variable chord with length 2 of this circle. Find the locus of the center of this chord.


By geometric thinking you can easily determine the locus. We will calculate the locus here in order to work with trigonometric functions and to illustrate a special elimination.

Take center of the circle in O (0.0).
Since the chord [AB] is as large as the radius, the included angle between OA and OB is pi/3 radians.
A(2.cos(t), 2.sin(t)) is a variable point of the circle.
Then, we have B(2.cos(t+pi/3), 2.sin(t+pi/3) )
The center M of the chord has coordinates

 
x = ( cos(t) +  cos(t+pi/3))
y = ( sin(t) +  sin(t+pi/3))
With Simpson's formulas ...
 
x = 2.cos(t+pi/6) . cos(pi/6)
y =  2.sin(t+pi/6) . cos(pi/6)
we simplify this to
 
x = sqrt(3).cos(t+pi/6)
y = sqrt(3).sin(t+pi/6)
These are actually the equations of two lines. These lines are the associated lines. They intersect in the middle of the variable chord.
 
     P(x,y) is a point of the locus
<=>

     There is a t-value such that

             x = sqrt(3).cos(t+pi/6)
             y = sqrt(3).sin(t+pi/6)

<=>
    the following system has a solution for t


             cos(t+pi/6) = x/ sqrt(3)
             sin(t+pi/6) = y/ sqrt(3)
                                     (elimination theory)
<=>

           x2/3 + y2/3 = 1

    <=>
           x2 + y2 = 3
The locus is a circle, as expected.

The equation of the locus arises by eliminating the parameter from the system formed by the associated lines.

Example a4

The points A (a, 0) and B (b, 0) are fixed points.
A point C is variable along the y-axis. Point D is the mirror image of C relative to the x-axis.
Find the locus of the intersection of AC en BD.
AC and BD are the associated lines


Take C(0,m), then D(0,-m). m is variable.
The equation of AC is mx + ay - am = 0.
The equation of BD is -mx + by + bm = 0.
 
    P(x,y) is intersection point of  AC and BD

<=>
   There is an m such that

     /  mx + ay - am = 0
     \ -mx + by + bm = 0

<=>
      The following system has a solution for m


      / (x - a)m = -ay
      \ (b - x)m = by
                                     (elimination theory)

<=>
      | x - a      -ay|
      | b - x       by|  = 0

<=>

     y = 0  or   x = 2 ab /(a + b)
The part y=0 arises at the moment that D = C. The most interesting part of the locus is the line x = 2 ab /(a + b).

The equation of the locus arises by eliminating the parameter from the system formed by the associated lines.

Example a5

Consider the hyperbola with equation y=1/x. Through a variable point P of the plane one takes a line parallel to the x-axis and a line parallel to the y-axis. These lines intersect with the hyperbola in the points A an B respectively. Find the locus of point P such that the centroid of the triangle PAB is on the line x=1.

Construct a figure with all the items on it. We choose variable coordinates (r,s) for P. Point P, which describes the locus, is the intersection point of the associated lines x=r and y=s. At first sight, there are two parameters r and s. But r and s are connected through a condition.

Point A is A(1/s,s) and point B is B(r,1/r). The centroid of the triangle PAB is Z((2r+1/s)/3 , (2s+1/r)/3). Z is on the line x=1 if and only if (2r+1/s)/3 = 1. This is equivalent to 2rs-3s+1=0. This is the condition that gives the connection between the parameters r and s. This equation 2rs-3s+1=0 is called a binding equation.

The equation of the locus arises by eliminating the parameter r an s from the system formed by the associated lines and the binding equation.

 
   x = r
   y = s
   2 r s - 3 s + 1 = 0.
We consider r and s as unknowns and we build the condition such that the system has a solution for r and s.
 
   r           =  x
   s           =  y
   2 r s - 3 s = -1
The condition is 2 x y - 3 y + 1 = 0. This is the equation of the locus. It can also be written in the form y = 1/(3-2x). The graph of the locus is a hyperbola.

Example 1 with parasitic parts

We start with a circle C with equation x2 + y2 = 1 and two fixed tangent lines p and q with equation x=1 and y=1.
A variable tangent line r to the circle intersects p in P and q in Q. Find the locus of the intersection point S of the line a through P parallel to the x-axis and the line b through Q parallel to the y-axis.

 
             
We take a variable point V(cos(t),sin(t)) on the circle. The tangent line in V to the circle is x.cos(t)+y.sin(t)=1.
The coordinates of P are (1, (1-cos(t))/sin(t))
The coordinates of Q are ( (1-sin(t))/cos(t) , 1)
The line a is y = (1-cos(t))/sin(t)
The line b is x = (1-sin(t))/cos(t)
The point S of the locus is the intersection point of the two associated lines a and b. The equation of the locus arises by eliminating the parameter t between the two equations of the associated lines.

We solve the system [x = (1-sin(t))/cos(t) ; y = (1-cos(t))/sin(t)] for sin(t) and cos(t).
sin(t) = (x - 1)/(x y - 1)
cos(t) = (y - 1)/(x y - 1)
Now we require that cos2(t) + sin2(t) = 1
After substitution, we bring all the non-zero terms to the left side and after factoring we obtain:
(x - 1) (y -1) (x y + x + y - 1)=0

The locus can be decomposed into three parts

Example 2 with parasitic parts

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   (3)

    (2)  <=>  y m2 + 2 m + y = 0   (4)
We eliminate m.
We ensure that m2 occurs only in one of the equations.
We calculate y.(3) - (x-2).(4) and after simplification, we have
 
      (x - 2) m2 + x = 0      (3)

      (x-2) m - y = 0          (5)
We now bring m from the second equation in the first equation. After simplification, we have
 
    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.

An alternative treatment

C is a circle with center A(1,0) and radius one and C' is a circle with center A'(-1,0) and radius one.

Through the origin O(0,0) we draw a variable line that intersects C' in a second point B'. Find the locus of the intersection point P of the line A'B' and the line through A orthogonal to the line OB'.

You can find the locus by giving the equation y=mx to the line OB'. Then you have to eliminate the parameter m between the equations of the lines AP and A'B'.

We will look into this problem with different approach. We start with a figure.

 
      

Let  Q be the intersection point of B'P and the circle C'.
OQ is orthogonal to OB' because  B'Q is a diameter of the circle C'.

Construct A'R perpendicular to OB'.
Now we have   A'R  //  QO  //  PA         (1)

Let S be the  intersection point of  AP and  OB'.
The triangle  ORA' is congruent with the triangle OSA.
So,   |O,R| = |O,S|          (2)

From (1) and (2)
   1 = |A',Q| = |Q,P|
So,  |A',P| = 2 = a fixed distance
The locus is a circle with center A' and radius 2.

Sometimes geometry is stronger than analysis. But the cooperation of both is brilliant!

Elimination of a quadratic parameter

The numbers a and c are constants and a > c > 0.
We start with two points F(c,0) and F'(-c,0) and we take a circle C with center F and a variable radius m. Now we take a second circle C' with center F' and a variable radius (2a - m).

The sum of the radii of C and C' is the constant number 2a.

We consider the two circles as associated curves and we'll calculate the locus of the intersection of the two curves.


C has equation (x - c)2 + y2 = m2
C' has equation (x + c)2 + y2 = (2a - m)2

We find the locus when we eliminate the parameter m from the system

 
 / (x - c)2 + y2 = m2
 \ (x + c)2 + y2 = 4 a2 - 4 a m + m2
But that parameter m appears quadratically in each equation of the system. Using the first equation we ensure that m2 disappears from the second equation.
 
 / (x - c)2 + y2 = m2
 \ (x + c)2 + y2 = 4 a2 - 4 a m + (x - c)2 + y2
We calculate m from the last equation.
 
   m = a - (c/a) x
We substitute this value in the first equation of the system.
 
  (x - c)2 + y2 =( a - (c/a) x )2
This is the equation of the locus, but we'll simplify this equation.
 
     (1- (c/a)2) x2 + y2 = a2 -c2

             Say  a2 - c2 = b2

<=>  (b2/a2) x2 + y2 = b2

<=>   x2/a2 + y2/b2 = 1

This curve is an ellipse. F and F 'are called the foci of the ellipse.

Note that a point P on the ellipse is if and only if | PF | + | PF | = 2a.

Hidden associated lines

The line d is a variable line through the fixed point A (a, 0) and that line d cuts the y-axis in point S. The perpendicular from S on d intersects the x-axis in point B.
Find the locus of the mirror image P of point B relative to the line d.

The line d has equation y = m(x-a). Here m is the parameter. Now we have
 
   S(0, -a m)

   Line  BS has as equation  y + a m = (-1/m) x

   Then we find   B(-am2, 0) and  P(a m2, -2 a m)
Now, we know the location of a variable point P of the locus, but we don't know the equation of the locus. At first glance, we see no associated lines.

But based on the coordinates of P, we can write:
The lines x = a m2 and y = -2 a m are intersecting in P and we can use these lines as associated lines.

Now it is sufficient to eliminate m from the system

 
   x = a m2
   y = -2 a m
One find a y2 = 4 x. That curve is a parabola with the x-axis as symmetry axis.

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.

Loci and parametric equations

Recall

The cycloid

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

Here is an illustration of the concept.

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.

Here is an illustration of the concept.

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


             


Example

The parametric equations of an astroid are x= cos3(t) y=sin3(t). Find the locus of the point P such that the tangent lines from P to the astroid are orthogonal.

First we take a variable point A(cos3(t), sin3(t)) on the astroid.

 
   The slope of the tangent line in A is

    dy     3 sin2(t) cos(t)
   ---- = ------------------  = - tan(t)
    dx     -3 cos2(t) sin(t)
Now, we take a variable point B(cos3(t'), sin3(t')) on the astroid.
 
   The slope of the tangent line in B is - tan(t')
The two tangent lines are orthogonal if and only if
 
         tan(t'). tan(t) = -1

   <=>   tan(t') = - 1/tan(t)

   <=>   tan(t') = tan(t+ pi/2)

   <=>     t' = t + pi/2  + k pi

   It is sufficient to study the cases
       t' = t + pi/2   en  t' = t + 3pi/2
   since the other values of t bring anything new.
The equation of the tangent line in A is
 
       y - sin3(t) = -tan(t) ( x - cos3(t))

<=>     y =  -tan(t).x + sin(t) cos2(t) + sin3(t)

<=>     y =  -tan(t).x + sin(t)         (1)

The equation of the tangent line in the point with t'= t + pi/2 is
 
        y =  -tan(t').x + sin(t')

<=>     y =  cot(t).x + cos(t)         (2)
The equation of the tangent line in the point with t'= t + 3pi/2 is
 
        y =  -tan(t').x + sin(t')

<=>     y =  cot(t).x - cos(t)         (3)
The tangent lines (2) and (3) are both orthogonal to the tangent (1).

First Case:

We start with the tangent lines (1) and (2)

 
         y =  -tan(t).x + sin(t)         (1)
         y =  cot(t).x + cos(t)        (2)
We calculate the point of intersection of these two tangents. We solve this system for x and y and we find after simplification
 
         x = (1/2) sin(2t) ( sin(t) - cos(t) )      (4)
         y = (1/2) sin(2t) ( sin(t) + cos(t) )      (5)
This gives us for each t the point from which the tangents to the asteroid are orthogonal lines.

(4) an (5) are the parametric equations of the locus. The the locus is indicated in blue.

 
    
Second Case:

We take the tangent (1) and (3). Now, one finds than for x and y

 
         x = (1/2) sin(2t) ( sin(t) + cos(t) )
         y = (1/2) sin(2t) ( cos(t) - sin(t) )
We'll show that this is the same locus as in the first case.

Since t is a real parameter, the shape of the locus does not change if we replace t by t + pi/2.

 
         x = (1/2) sin(2t+pi) ( sin(t + pi/2) + cos(t + pi/2) )
         y = (1/2) sin(2t+pi) ( cos(t + pi/2) - sin(t + pi/2) )

<=>      x = -(1/2) sin(2t) ( cos(t) - sin(t) )
         y = -(1/2) sin(2t) ( -sin(t) - cos(t) )

<=>      x = (1/2) sin(2t) ( sin(t) - cos(t) )
         y = (1/2) sin(2t) ( sin(t) + cos(t) )
Now, it is clear that this locus is the same one as in the first case.

Note:

The locus is tangent to the asteroid in eight points. Let C be such a point. The two tangent lines from C to the asteroid are orthogonal lines. Since C is on the asteroid the first tangent is the tangent in C to the astroid. The other tangent is a normal line in C.
Conclusion:

The eight special points are the only points of the astroid such that the normal line in these points are also tangent lines to the astroid.

Exercises

The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the problem before you read the given solution.




Topics and Problems

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