1

Locus Problems With Complex Numbers

Locus

We start with a definition concept of locus. Then, we will present some basic examples of

locus problems, given as regions in the complex plane.

A locus is a set of points whose members are determined by a specific rule.

This set of points, when marked on a coordinate system, may look like a line, curve or a

surface. Loci can be given in equation forms which involve complex numbers.

Some Basic Examples

Ex 1 Let z be a complex number. Then the set of points which is defined as

{( , ) 5Re( ) },| zL x y z C

is an example of a locus.

For brevity, we can denote this locus by the equation

5 Re( ).z z

From now on, we will use equation forma to denote a loci.

We need to see, how this set of points appear, when we plot them on a Cartesian plane. To

explore this, let .z x iy Then,

2 2 2 2 2 2 2

This restricts the values of . Why and how?

55Re( ) 5 4 2 (Eq.1)

x

x y xx iy x iy x y x y x y x

2

Is the locus of Ex 1, really 2 ?y x Which one of the following two graphs depicts the

locus of 5 Re( )z z ?

Graph I: 2y x

Graph II: 2 , 0y x x

To answer this question, think of the question raised in the red-comment in (Eq.1):

2 2

The condition in the This quantity is LHS, demands that 0.always 0. Why?

5

x

x y x

Graph I admits values of 0,x and Graph II does not. This will help you to choose the

correct picture of the locus.

3

Consider the loci of the following equations.

Ex 2. What is the shape of the locus given by the equation, ?5 Re( )z z

2 2 .5 Re( ) 5x y xz z

The reader can see that the restriction marked in the (Eq.1), is no longer valid. Therefore, the

correct graph of the locus is given by Graph I.

From now on, we will use equation form to denote a locus.

Ex 3 Sketch the locus given by the equation, 2 3

Re( ) Im( ) 64.z z

Let .z x iy Then

32

2 3Re( ) Im( ) 64 Re( ) Im( ) 64.

x y

z z z z

Therefore, the equation yields

2 3 64. (Eq.2)x y

i. Can you notice that (Eq.2) limits the domain of y? Why is this?

ii. The reason is based on Year 7 or Year 8 Mathematics. Namely, x , being a real number, 2 0.x Therefore,

iii 3 64 4.y y

iv However, there is no restriction for the values of x , since for any value of x , there is a

matching value of y .

v When 4 0.y x

4

vi Now, start the graph, beginning from the point, (0,4). When the magnitude of the variable,

x, is increasing then magnitude of the variable, y, has to decrease, so that (Eq. 2)

continues to be satisfied. vii When 0,x y remains closer to zero. This should happen for a long interval of x since

3

much muchsmaller

x x , for 0.x

(viii) Therefore, for a longer interval of x near zero, the change in y is small. Therefore top of

the graph should look flat.

(ix) Also, when 8, 0.x y

(x) When 8x or 8x , the value of 0.y This means that near x = 8 and x = -8. The

graph should look almost vertical.

(xi) Since 3 22 3 64 64x y y x , for values of x with very large magnitude,

1 12 23 3 .( ) ( )x xy This means that as x gets larger and larger in magnitude y is tending

to be a larger and larger negative number.

(xii) Moreover, for large values of x, the graph of 1 2

2 3 3-( ) -x xy . This is because when

the variable x has a large magnitude, 2 2.64 xx To convince yourself about this, use

your calculator to compare the values of

3 3 3 3;64 1064 and 1064 64 20000 and 20000 .

(There was no magic in my choice of the values for the variable, x. I just needed some

reasonably large values.)

Now we present the sketch of the locus.

5

Can you figure out how one can use the features, (i)-(xii), to produce the graph above,

without using a graphing calculator? After you figure it out, you may check this graph

using your calculator.

Ex 4 Sketch the locus given by the equation, 2

.z z

2

2

This equtionresembles an equation such ast where .

So, we can factorise out. The Null

0 unit circleFactor Law

( 1) 0 0 or 1

t t Z

Z z

z z z z z z

So the required locus is the unit circle and its centre. If one does not include the point (0,0) in

the locus, the solution is incomplete. The answer requires that you mark the point, (0,0), along

with the circle.

6

Ex 5 Sketch the locus given by the equation, 2 .Z Z

Let .z x iy Then,

2

2 2 2 2

In this side, Im Part =0 and Re Part 0

2 .

z z

x y xyi x y

The imaginary part in the RHS (Right Hand Side) is zero. Therefore, the imaginary part of the

LHS must also be zero. Hence, we have

2 0 0, 0.xy x y or

This gives the equation:

2 2 2 2

This value 0

. (Eq.*)x y x y

Since 2 2 0,x y if x = 0, (Eq. *) gives

.

2 2

This is not or y, but

.

y

y

y y y

Similarly, if y = 0, (Eq. *) gives

2 22 .x y xx

Then, the condition, x = 0 gives that

2

a negative number or positive number or zero. zero.

0 (0,0).y zy y

7

And, the condition, y = 0, gives that

when is nonnegative

when is non-positive

2

2

2

1, or 0 (0,0), (1,0) 0 1,

1, or 0 (0,0), ( 1,0) 0 1.

x

x

x x z

x x z

x

x x

x

or or

or or

Then, the locus of 2z z is

.{ 0, 1, 1}z z z

The answer requires that you mark these points on the plane.

Ex 6 Sketch the locus given by the equation, 2 2( 1) 1.z z

There is an almost self-harming way to proceed with this. With ,z x iy we can write

8

22 2 2 2 2

2 22 2 2 2 2 2 2 .

( 1) 1 ( 1) 2( 1) 1 2

( 1) 4( 1) 1 4

x iy x iy x y x iy x y xyi

x y x x y x y

To solve this, first, one has to simplify the contents inside the square roots, square both sides,

and then solve the resulting equation. You can see how humongous the task is, but there is an

easier way.

2

2

( 1)( 1)

1( 1)( 1)

2 22

2

1

Then,

0 0

( 1) 1 ( 1) ( 1) 1 ( 1)( 1) 0.

1 1 1 ( 1) 1 1 .

z z

zz z

zz z z z z z z

z z z z z z

At this point, the humble Null Factor Law comes to our rescue:

Absolute value of a number is zero iffthe number is zero.

Hence, 1.

0 0 or 1 1( 1) 1 1 ( 1)

z

z zz z z z

Accordingly, one of the points in the locus is 1.z To get the other points, we need to solve

the second equation, 1 1z z . With ,z x iy we get

2 2 2 2 2 2 2 2( 1) ( 1) ( 1) ( 1)

2 2 0.

1 1 x y x y x y x y

x x x

z z

Finally, the answer requires that you mark the point 1,z and the whole y-axis.

9

Hyperbolas

Hyperbolas can be described by the following principle:

In the diagram above, A and B are two given points, P is a variable point on the

Cartesian plane, and k is a positive number. Then the locus,

AP BP k

is a hyperbola provided the number k satisfies a necessary condition.

To further investigate this condition on k, first we refer to the triangle inequality (in the

geometric form): Length of any side of the triangle must be smaller or equal to the sum of the

lengths of the other two sides. Therefore, we can write that

,AB BP AP

(1,0) x

y

10

where the equality holds if and only if the point B lies in the straight line segment AP between

A and P. Now, we rewrite the inequality in the following form:

.AB BP AP AP BP AB

Should this condition hold, even if the quantity, AP-BP is positive? To explore this, we also

note that we can write another triangle inequality for the same triangle:

,BA AP BP

where the equality holds if and only if the point A lies in the straight line segment BP between

B and P. Then,

.BA AP BP BP AP AB

Therefore both conditions AP BP AB and BP AP AB must hold. This can be rewritten

in a more compact form as

. (Eq.3)AP BP AB

Therefore, it is clear that one cannot find points P such that AP BP AB . And, if

AP BP AB

Then, either or .AP BP AB AP BP AB Now, observe that

is a straight line.AP BP AB AP AB BP ABP

On the other hand, if AP BP AB , then we have

is a straight line.AP BP AB BA AP BP BAP

11

Examples

The discussion above is applicable to the following locus problems. They could be hyperbolas,

depending on quantity in the RHS.

a) 2 4 3 4 4z i z i

b) 2 4 3 4 5,z i z i

c) 2 4 3 4 6.z i z i

Each of these locus problems involve 3 points in the Cartesian plane:

P: z x iy

A: 2 4v i

B: 3 4w i

2 4z iAP z v

3 4BP z i

2 4 3 4 5 5AB v w i i

12

Note that the equation,

2 4 3 4 4z i z i ,

can be written as

  4 4,AP BP AP BP

but

(Eq.3)

4 5.AP BP AB

Therefore, the equation, 2 4 3 4 4z i z i should result in a part of a hyperbola. If

the Example (a) is changed to

,2 4 3 4 4z i z i

then the result would be a complete hyperbola.

Moreover, the locus,

2 4 3 4z i z i p ,

will be empty for any 5;p for 5p the locus will be a part of straight line, and for 5,p

the locus will be a hyperbola.

Detailed Solution: Example (a)

To find out the exact locus of 2 4 3 4 4,z i z i first we let ,z x iy and then write:

2 4 3 4 4 2 4 3 4 4.z i z i x iy i x iy i

This will lead to

13

2 2 2 2

Students should learn to come to this step, directly from

2 4 3 4 .

4

4 ( 2) ( 4) ( 3) ( 4)

2 4 3 4 4 2 ( 4) ( 3) ( 4)

2 4 3 4

z i z i

x y x y

x iy i x iy i x i y x i y

x iy i x iy i

.4

Now, move the second term in the LHS of the equation,

2 2 2 2( 2) ( 4) ( 3) ( 4) 4,x y x y

to its RHS, and then square both sides to get:

22

2 2

2

2

2 22 .

4 ( 3) ( 4)

=

( 2)

16 + 8 ( 3) ( 4)

(

( 3)

4)

( 4)x

x yx

x

y

yy

Now, we cancel out the red terms in the equation above. Next, we isolate the square root

term, by obtaining the following equation:

2 2 22 - 16 8 (( = 3) ( 4)( 2) 3) x yx x

By simplifying the LHS, we get,

21

2

10

2

must be nonnegative since the RHS is nonnegative:

10 21 8 ( 3) ( 4) . (Eq.4)

x

x x y

Now square both sides to get,

2 2 210 21 64( 3) 64( 4) .x x y

14

Then, by moving the right hand term to the LHS and then simplifying and completing square

on x terms, we get,

2

2136 64( 4) 144.

2x y

At this point, by dividing both sides by 144, and then writing the result in the standard form,

we get:

From(E

2

q.4)

2

2

2

( 4

12

12

211

)and .

32

0

xy

x

The hyperbola exists only when (to see this apply the principle, if 2 2 21 1.a b a )

2

2

1 1 12 2 2

1 1 or 12 22

,

x x x

.

These inequalities yield or .3 5

2 2

x x One of these two inequalities should be

satisfied together with our early inequality, 2110

x . That is,

.No points 3for 2

Impossible 52

. (Eq.5)and oror 3 5 3 5 2 2 2 2

21 21 2110 10 10

x

x

x x x xx x x

and and

The overall condition, ,52

x gives the RHS of the hyperbola.

15

The following sketch depicts the sketch of the hyperbola,

2

2

2 2

12 ( 4)

12 3

2

xy

.

If you provide the sketch above as your answer for the locus, you will loose marks. The

following is the correct locus is given in brown in the sketch below.

16

Here is the graph of the solutions for our original complex equation.

What would have happened, if we had 2 4 3 4 4?z i z i

2 2 2 24 ( 2) ( 4) ( 4 ( 3) ( 4) )2 4 3 4 x y x yz i z i

22 2222( 16 8 (( 42) ( 33)) )4) 4)( (x y xy yx

2 2 2 2( 2) ( 3) 16 8 ( 3) ( 4) .x yx x

Finally, we obtain

2 2 (Eq.6)10 21 8 ( 3) ( 4) .x x y

We could have obtained (Eq. 6), just by observation of the work we did to obtain (Eq. 4),

without repeating all the work. Also, the only difference of this equation and (Eq.4) and

(Eq. 6) is the presence of the negative sign before the square root in the RHS of the latter

equation.

Now, accordingly, the inequalities in (Eq. 5) must be modified:

No points for 5.2

3 Impossible2

. (Eq.7)and and or andor 3 5 3 5 2 2 2 2

21 21 2110 10 10

x

x

x x x xx x x

Hence, this new locus gives the, same hyperbola for 32

x , which is LHS of the hyperbola.

Consequently, the locus of

2 4 3 4 4z i z i

is the complete hyperbola as both inequalities of (Eq.5) and (Eq. 7) are satisfied.

17

Example (b)

We now work with the second equation, 2 4 3 4 5:z i z i

Note that this can be written as - 5.AP BP As we have discussed earlier, if AP-BP=AB,

this equation must contain points lying somewhere on AB. Since AB = 5 and .AP BP AB

we have the following diagram.

Can you guess the solution set from the diagram? The solution set must be

{( ,4): 3}.x x

This is because AP > AB and the point, P, lies on AB.

Let us see, whether we can reach the same solution, by further analysis,

Now,

2 4 3 4 5 2 4 3 4 5.z i z i x iy i x iy i

This final equation above, is equivalent to the equation:

2 2 2 2( 2) ( 4) ( 3) ( 4) 5x y x y

Now, move the second term to the left hand side and square both sides of the resulting

equation:

2

2 2 2 2

2 2 2 2.

( 2) ( 4) 5 ( 3) ( 4)

= 25 + 10 ( 3) ( 4) ( 3) ( 4)

x y x y

x y x y

As a result, we get

18

2 2 2 2 .( 2) ( 3) - 25 = 10 ( 3) ( 4)x x x y

By simplifying we get,

2 210( 3) 10 ( 3) ( 4) .x x y

Since the RHS of the equation above is nonnegative, we have 3.x

Now, after dividing by 10, and then squaring both sides, we obtain:

2

2 23 ( 3) ( 4) .x x y

This leads to the equation:

2( 4) 0 4y y

Now, having established that always 4,y let us investigate further whether we can also

verify that 3,x employing this kind of analysis.

Since y = 4, now the original equation, 2 4 3 4 5x iy i x iy i reduces to

4 2 4 4 3 4 2 3 5.x i i x i i x x

For 3,x the condition, 2 3 5,x x further reduces to

2 ( 3) 5.x x

The left hand side always simplified to 5. This indicates that the original equation,

2 4 3 4 5x iy i x iy i , is valid for any x satisfying the constraint 3,x given that

y = 4. Note that this solution interval is on the right hand side of the complex number v on

the straight line joining w and v. Next, we consider another possible interval of solutions

for x.

19

For 2 3,x the condition, 2 3 5x x , becomes

2 ( 3) 2 ( 3) 2 1 5 3.x x x x x x

This indicates that the original equation, 2 4 3 4 5,x iy i x iy i is valid for x =

3, in the interval [ 2,3],x but this solution is already covered in the case that 3.x

Finally, consider the interval ( , 2).x Then the condition, 2 3 5x x , reduces to

2 ( 3) 5 5.x x

This absurdity indicates that the original equation (the second equation) has no solutions for

( , 2).x Therefore, the complex numbers, z, satisfying the Example (b), lie on the right

hand side of the straight line joining w and v, including the complex number v.

Thus the locus of Example (b) is given in the following sketch.

Here AP-BP =AB, as long as the point P on the line AB, to the right of the point B.

Now consider Example (c).

2 4 3 4 6x iy i x iy i

Then

2 2 2 2( 2) ( 4) ( 3) ( 4) 6x y x y

20

Now, move the second term to the LHS and square both sides of the resulting equation to

get:

22 2 2 2

2 2 2 2

( 2) ( 4) 6 ( 3) ( 4)

= 36 + 12 ( 3) ( 4) ( 3) ( 4)

x y x y

x y x y

So, we get

2 2 2 2( 2) ( 3) - 36 = 12 ( 3) ( 4)x x x y

By simplifying we get,

2 210 41 12 ( 3) ( 4) .x x y

Since the RHS is nonnegative we should have:

41

. (Eq.8)4

x

Now square both sides to obtain:

2

2 210 41 144( 3) 144( 4) .x x y

That is,

2 2 2100 820 1681 144( 6 9) 144( 4) .x x x x y

Now, by moving the first two terms of the LHS to the RHS, and then simplifying and

completing square on x terms, and then finally flipping the sides, we get,

2

2144 144( 4) 396.

2x y

Next, by dividing both sides by 396, and then by writing the result in the standard form, we

get,

21

2

2

2 2

12 ( 4)

1.3

112

xy

Therefore each of the term in the LHS should be less than or equal to 1. Thus, from the first

term of the LHS, we have:

2

2

12 5 7 1 .

2 23

x

x

Since it is required by (Eq. 8) that41

4

x , the original complex number equation does not

have any solution, and consequently the graph is empty.

Also we may say that since the original equation indicates a hyperbola and its

simplified form is an equation an ellipse there are no complex numbers z satisfying the

equation, 2 4 3 4 6.z i z i

Ellipses

Examples

(A) 2 4 3 4 4z i z i

(B) 2 4 3 4 5z i z i

(C) 2 4 3 4 6z i z i

From a well-known definition of an ellipse, it follows that the respective simplified equation

of each of the equations must yield a possible equation of an ellipse.

22

Note that the distance between w and v is 5 units. So, triangle inequality imposes that

. AP BP AB

As we have already mentioned before, since complex numbers can be identified with

appropriate points in the plane, this triangular inequality dictates nature of the loci of the

examples above.

Since, we know that AB = 5, the locus of the relationship,

2 4 3 4 .

PBAP

z i z i p

is a null set if p < 5. Again, this is because, any three points on the plane should abide by

the triangular inequality.

Now, let us consider the equation in Example (A).

2 4 3 4 4.z i z i

Since, this implies that

5

AP BP AB ,

there are no solutions. That is, the solution set is empty. However, as a mathematical

expedition, we carry out further analysis.

23

First, note that

2 4 3 4 4 2 4 3 4 4z i z i x iy i x iy i

Then,

2 2 2 2 .

2 4 3 4 2 ( 4) ( 3) ( 4)

= ( 2) ( 4) ( 3) ( 4)

x iy i x iy i x i y x i y

x y x y

Thus,

2 2 2 2( 2) ( 4) ( 3) ( 4) 4. (Eq.9)x y x y

Each of the quantities in the LHS of the equation above is nonnegative. Hence, each of them

must be less than or equal to 4. We can write this in the form:

2 2

2 2 2 2 2 2

( 4) 0 ( 4) 0.

( 2) ( 4) 16 ( 3) ( 4) 16 ( 2) 16 ( 3) 16.

y y

x y x y x x

and and

That is,

2 2( 2) 16 ( 3) 16.x x and

Even though a similar inequality, 2( 4) 16 0 8,y y can be derived for y, it is not

useful at the moment. The first of the two aforementioned inequalities on x, gives that

4 ( 2) 4 6 2 .x x

And, then, the second inequality, indicates that:

4 ( 3) 4 1 7.x x

Again, both of these equations are dictates of the initial equation. Therefore, the initial

equation has solutions only for

24

1 2. (Eq.10)x

Now, move the second term of the LHS of (Eq. 9), to the RHS and square both sides of the

resulting equation, to obtain:

22 2

2

22

2 2 2

4 ( 3) ( 4)

= 16 - 8 ( 3)

( 2)

(

( 4)

( 4)34) ( )

x y

x y

x

x

y

y

Consequently, we get:

2 2 2 2( 2) ( 3) - 16 = 8 ( 3) ( 4)x x x y

By simplifying, we arrive at:

2 210 21 -8 ( 3) ( 4) . (Eq.11)x x y

Since, the LHS is always non-positive, we have

21. (Eq.12)

10x

Now, square both sides of (Eq. 11), to get,

2

2 210 21 64( 3) 64( 4) .x x y

Next, move the RHS to the LHS of the equation above, then simplify, and complete square

on x terms, to obtain,

2

2136 64( 4) 144.

2x y

Dividing both sides by 144, and then writing in the standard form we receive that,

25

From this equation, it is evident that

2

2

12 1 1

1 3 and 3.2 23

x

x x

This yields the inequalities,

.5 7

and (Eq.13)2 2

x x

For our original complex equation to have solutions, all the inequalities given by (Eq.10),

(Eq.12), and (Eq.13) must be obeyed.

That is, if explicitly said, all the inequalities, 2,1 x 7,

2x

21,

10x and

5,

2x must

be obeyed. It is obvious that no values of x can satisfy all of these inequalities.

Therefore, the locus of Example (A) is an empty set.

Also, since the original equation indicates an ellipse, and its worked out

form is an equation of a hyperbola we may conclude that there are no

complex numbers, z satisfying the equation,

2 4 3 4 4.z i z i

2

2

2 2

142

1.3 3

2

xy

26

We now work with Example (B).

The equation gives

52 4 3 4 5AB

PBAP

AP PBx iy i x iy i

This can be pictured in the following diagram:

It is clear that P must lie between A and B. This gives the inequality, -2 x 3. Therefore,

the solution set of Example (B) must be

{( ,4): 2 3}. (#)x x

For the moment, we forget this, and will conduct all the calculations, to see, how the

mathematics is being played out.

We start with the equation

2 2 2 2( 2) ( 4) ( 3) ( 4) 5x y x y

Now, move the second term of 2 2 2 2( 2) ( 4) ( 3) ( 4) 5x y x y to the RHS and

square both sides of the resulting equation, to get:

27

22 2 2 2

2 2 2 2

( 2) ( 4) 5 ( 3) ( 4)

= 25 - 10 ( 3) ( 4) ( 3) ( 4)

x y x y

x y x y

Thus, we get

2 2 2 2( 2) ( 3) - 25 = 10 ( 3) ( 4)x x x y

By simplifying we obtain,

2 210( 3) 10 ( 3) ( 4) .x x y

Now, after dividing by 10 and squaring both sides of the resulting equation, we arrive at:

2

2 23 ( 3) ( 4) .x x y

This leads to the condition:

2( 4) 0 4.y y

This means that all the solutions for Example B, must have just single y value, y= 4. Since

y = 4, the equation

2 2 2 2( 2) ( 4) ( 3) ( 4) 5x y x y

indicates that (since each of the terms in the LHS must be less than or equal to 5)

2 2

2 2 2 2 2 2

4 ( 4) 0 4 ( 4) 0

( 2) ( 4) 25 and ( 3) ( 4) 25 ( 2) 25 and ( 3) 25.

y y y y

x y x y x x

This equation yields two inequalities,

5 2 5 and -5 3 5, and 4.x x y

Consequently, we conclude that

28

7 3 2 8 and 4 7 2 and 4.x x y x y and

2 3 and 4. (Eq. 14)x y

The equation, (Eq.14) gives the only feasible solution for Example B.

Just to double check this, let us revisit the very original form of Example (B).

Since the condition, y = 4 must be satisfied with any possible solutions for x we start by

substituting, as follows:

2 4 3 4 5 4 2 4 4 3 4 5

2 3 5.

x yi i x yi i x i i x i i

x x

For 3,x the equation, 2 3 5x x , can be written as:

2 ( 3) 2 1 5 3.x x x x

This indicates that Example (B), in the interval of [3, ), has only one solution, 3.x

In the interval of[ 2, 3] , the equation, 2 3 5x x , becomes

2 ( 3) 2 ( 3) 5.x x x x

This indicates that the original equation, Example (B), is satisfied for all the values of x in

the interval, 2 3 4.x y and See again (#) and (Eq. 14).

Finally, consider the interval of ( , 2). Here, the LHS of the equation, 2 3 5,x x

reduces to

29

2 ( 3) 5.x x

Then, we have,

-5=5.

This absurdity indicates that the original equation, Example (B), has no solutions in the

interval of ( , 2). Therefore, the set of complex numbers, z, that satisfy the Example

(B) lie on the straight line segment that joins v and w. The complex numbers v and w are also

in the locus. This is another verification of the already obtained result; see (Eq. 14).

Next, we consider the Example (C):

2 4 3 4 6.x iy i x iy i

Then, by setting ,z x yi we obtain that

2 2 2 2( 2) ( 4) ( 3) ( 4) 6.x y x y

Now, move the second term to the LHS and square both sides of the resulting equation:

22 2

22

2 2

22

6 ( 3) ( 4)

= 36 + 12 ( 3)

( 2)

(

( 4)

( 4)34) ( )

x y

x y

x y

x y

This leads to the equation,

2 2 2 2 .- 36 =( 2) ( 3 12 ( 3 ( 4)) )xx x y

By simplifying we get,

2 210 41 -12 ( 3) ( 4) . (Eq.15)x x y

Since the LHS is always non-positive, we have the condition that

30

41. (Eq.16)

10x

Now square both sides of (Eq. 15), to get:

2

2 210 41 144( 3) 144( 4) .x x y

That is,

2 2 2100 820 1681 144( 6 9) 144( 4) .x x x x y

Now move the first two terms of the LHS of the equation above, to the RHS, simplify,

complete square on x terms, and finally flip the sides, to get,

2

2144 144( 4) 396. (Eq.17)

2x y

Since, each one of the two terms in the LHS is nonnegative, none of those terms can be larger

than 396. Therefore, we get the inequality:

2 2

13.

21 144 396 9 32 2

xx x

This gives the inequality:

. (Eq.18)5 72 2

x

Now, by dividing both sides of (Eq. 17) by 396, and writing the resulting equation in the

standard form, we get,

31

2

2

2 2Eq.19

12 ( 4)

1.3

112

xy

Now the inequality, 4110

x (Eq. 16) is satisfied by the inequality, 5 7

2 2x

, (Eq. 18).

This means that the whole of the domain of (Eq. 19), (or equivalently, (Eq.17)), is allowed

by (Eq. 16). Thus the locus is the whole ellipse given by (Eq. 19).

The ellipse shown below, is the required locus.

The condition on y,11 11

4 42 2

y , which can be seen on the diagram above, can be

obtained from (Eq. 17), since the largest value possible for the second term in the LHS of

(Eq. 17) is 396. You may also refer to the derivation of the inequality in (Eq. 18).

32

In this article, I have repeated some information, even several times, within

the same problem. This shouldn’t be necessary. However, I adopted this style

because it is my experience that many students are unable to remember the

first sentence after reading the second sentence, while following a

mathematical argument.

This defect of many students may be due to playing of video games in which

what matters is only the current screen, not anything before or anything after.

In contrast, mathematics consists of trains of thoughts. If a link is broken

somewhere, the rest and the whole becomes incomprehensible.