Chapter 7: Eigenvalues and Eigenvectors 16

SECTION G Sketching Conics

By the end of this section you will be able to

 recognise equations of different types of conics
 complete the square and use this to sketch a given conic
 use the theory of eigenvalues to sketch conics with cross - product terms

In this section you will need to know your work from previous sections. Throughout this
section we will assume that you can find eigenvalues, eigenvectors, orthogonal diagonalize a
symmetric matrix and convert quadratic forms xT Ax into diagonal form.
In addition you will need to know how to complete the square. Completing the square is not
covered in this book so you need to familiarise yourself with this before covering this
section. You will also need to know the basic equations and shapes of a circle, ellipse,
parabola and hyperbola. Please read up these two topics if you are not familiar.
G1 Definition of a Standard Conic
What does the term conic mean?
A conic is a shape obtained by a plane cutting a double cone at various places as illustrated
below:

Figure 2 (a) (b) (c) (d)

Circle Ellipse Parabola Hyperbola
Depending on where the plane cuts the cone we have the above conic shapes – circle, ellipse,
parabola and hyperbola. This was first developed by the Greek mathematician Apollonius
262BC to 192 BC. At that time there was no algebraic interpretation of these conics because
the marriage between algebra and geometry did not take place until Descartes coordinate
system was developed in the 16th century.
Conic sections describe the orbits of planets. Before Descartes developed his coordinate
system, the above geometric interpretation was how conic sections were described. Newton
in his Mathematical Principles of Natural Philosophy used the above geometric idea to
describe the motion of planets. He found it difficult to follow Descartes work and there was
no expert at Cambridge who could help him digest Descartes analytic geometry.
The basic conic shape can be drawn on Cartesian coordinate system with its equations as
follows:
x2 y 2
x2  y 2  a2  1
a 2 b2

(a) Circle (b) Ellipse

Chapter 7: Eigenvalues and Eigenvectors 17

b b
y x y x
a a
x2 y 2
 1
a 2 b2

Figure 3 (c) Parabola with a  0 (d) Hyperbola

What do you notice about the equations for the above conics?
They are all of quadratic form in two variables x and y. The general equations of various
conics in standard form are given by:
(7.45) (a) x 2  y 2  a 2 is a circle with centre origin (0, 0) and radius a.
x2 y 2
(b)   1 is an ellipse with centre origin (0, 0) and crosses the x-axis at a and
a 2 b2
a and y-axis at b and b .
(c) y  ax 2 is a parabola.
x2 y 2
(d) 2  2  1 is a hyperbola with centre origin (0, 0). There are two lines that the
a b
b b
hyperbola comes closer and closer to (asymptote) which are given by y   x and y  x .
a a
The above formulae in (7.45) (a), (b), (c) and (d) is the standard form for conics.
In this section you will need to identify the graph of each of these equations and sketch them
on a Cartesian coordinate system as the next example illustrates. Also we will only
concentrate on ellipses, hyperbolae and circles.
Example 48
Sketch the following graphs:
(a) x 2  y 2  4 (b) x 2  4 y 2  36 (c) x 2  5 y 2  25
Solution
(a) What type of graph is given by the equation x 2  y 2  4 ?
x 2  y 2  4  22 is the equation of a circle with centre origin (0, 0) and radius 2. We have

x2  y 2  4

Figure 4
(b) We need to place the given quadratic x 2  4 y 2  36 in standard form. How?
Divide through by 36. We have
x2 4 y2 x2 y2 x2 y2
  1 implies   1 implies 2   1
36 36 62 9 6 32
x2 y2
What graph is represented by this equation 2   1?
6 32
Chapter 7: Eigenvalues and Eigenvectors 18

This is a hyperbola because the standard form of a hyperbola is given by the formula
x2 y2 x2 y2
  1 . In this case,   1 , what are the values of a and b?
a2 b2 62 32
a  6 and b  3 so the graph cuts the x axis at 6 and 6.
b 3 1 b 1
Also the hyperbola becomes closer and closer to y  x  x  x and y   x   x :
a 6 2 a 2

1
x 2  4 y 2  36 y x
2

1
y x
2

Figure 5
x2 y2
Remember the given equation x  4 y  36 is equivalent to 2 
2 2
 1.
6 32
(c) The given equation x 2  5 y 2  25 is not one of the above conic formats in (7.45). How do
we convert this into a standard format of a conic?
Dividing the given equation x 2  5 y 2  25 by 25 yields
x2 y2 x2 y2 x2 y2
 5  1 simplifies to   1 which gives 2   1
 
2
25 25 25 5 5 5
x2 y2
What type of conic is given by this equation   1?
 
2
52 5
x2 y2
Ellipse because the general formula of an ellipse is   1 and it crosses the x-
a2 b2
axis at  a and y-axis at b with centre origin (0, 0). What are the values of a and b in our
case?
x2 y2
a  5 and b  5 . The graph of x 2  5 y 2  25 is equivalent to 2   1 which
 
2
5 5
is an ellipse that crosses the x-axis at 5 and 5 and y axis at 5 and  5 .

Figure 6
Chapter 7: Eigenvalues and Eigenvectors 19

G2 Conics Not in Standard Form

In the above Example 48 the conic was either in standard form or could easily be placed into
standard form by dividing through by a factor. However it is more difficult to sketch graphs
of quadratics which are not in standard form. For example how do we sketch the quadratic
x 2  8 x  4 y 2  24 y  36 ?
We need to convert these into standard form by completing the square. As mentioned at
the start of this section you will need to be familiar with completing the square.
Example 49
Write x 2  8 x  4 y 2  24 y  36 in standard form and sketch the graph.
Solution
We first need to complete the square on x 2  8 x  4 y 2  24 y  36 . We
complete the square on x and y variables separately. First we examine the x variable:
x 2  8 x   x  4   42
2
[Completing the square on x terms]
Considering the y variable in the given equation:
4 y 2  24 y  4  y 2  6 y  Taking out 4
 4  y  3  32  Completing the square on y terms
2
 
 4  y  3  36
2

Substituting these x 2  8 x   x  4   42 and 4 y 2  24 y  4  y  3  36 into the given

2 2

equation x 2  8x  4 y 2  24 y  36 yields

 x  4  16  4  y  3  36  36
2 2

 x  4
 4  y  3  16
2 2

If you found the above derivation difficult to follow then you will need to go over completing
the square and then try to digest the above. However this  x  4   4  y  3  16 is
2 2

still not in standard form for any of conics in (7.45). How do we convert this into one of those
formats?
Let new variables x’ (x dashed) and y’ (y dashed) be defined by x  x  4 and y  y  3 .
From above we have
 x  4   4  y  3  16
2 2

 x   4  y   16 Substituting x  x  4 and y  y  3
2 2

 x   4  y   1
2 2

2 2
 Dividing through by 16  42 
4 4
 x   y 
2 2

2
 2
1  Because 4  22 
4 2
 x   y 
2 2

We say x’ and y’ are our new axes. How do we sketch the graph of   1?
42 22
 x   y   x   y 
2 2 2 2

 1 is in standard form and fits the equation of the ellipse 1 

42 22 a2 b2
which is based on the axes x and y’. Where is the origin of this new axes, x and y’, with
reference to our normal x and y axes?
Chapter 7: Eigenvalues and Eigenvectors 20

In the above we defined the new variables (axes) as x  x  4 and y  y  3 which means
the origin in this new system is where x  0 and y  0 . This occurs when x  4  0 and
y  3  0 which gives the point x  4 and y  3 respectively. What does this mean?
Means that the origin of the new system is 4 units left and 3 units up from our normal origin.
 x   y 
2 2

Hence the ellipse   1 is centred at  4, 3 with length 4 along each arm
42 22
of the x ' axis and length 2 along each arm of the y ' axis as shown below in Fig. 7.

2
4

Figure 7
 x   y 
2 2

Note that x 2
 8x  4 y 2
 24 y  36 is equivalent to   1.
42 22
By examining the given equation x 2  8 x  4 y 2  24 y  36 in the above
example we could not say it is an ellipse but once converted into the new axes we can
immediately see that we have an ellipse.

Example 50
Write the quadratic x 2  4 x  y 2  6 y  14 in standard form and sketch the
graph.
Solution
Again we do not know what shape the graph of the given quadratic is but by converting into
another format we may be able to conclude it is one of the conics given in formula (7.45).
We first need to complete the square on x 2  4 x  y 2  6 y  14 . Completing
the square on each variable separately:
x 2  4 x   x  2   22
2
[Completing the square on x terms]
Consider the y variable:
 y2  6 y    y2  6 y  Taking out the minus sign 
   y  3  32  Completing the square on y terms 
2
 
   y  3  9 Taking in the minus sign 
2

Substituting these x 2  4 x   x  2   22 and  y 2  6 y    y  3  9 into the given

2 2

equation x 2  4x  y2  6 y  14 yields:
 x  2  22   y  3  9  14
2 2

 x  2    y  3 9  Re-arranging 
2 2

Again this  x  2    y  3  9 is not in standard form of a conic. How do we convert this

2 2

into one of the conic formulae of (7.45)?

Chapter 7: Eigenvalues and Eigenvectors 21

Let new variables x ' (x dashed) and y ' (y dashed) be defined by x  x  2 and y  y  3 ,
then substituting these into the above  x  2    y  3  9 gives
2 2

 x  2    y  3  9
2 2

 x    y   9 Substituting x  x  2 and y  y  3
2 2

 x    y    1
2 2

 Dividing by 9  32 
 
32 32
 x   y 
2 2

How do we sketch the graph of 2

 1?
3 32
 x   y   x   y 
2 2 2 2

 1 is in standard form and is an equation of the hyperbola 1 

32 32 a2 b2
which is based on the axes x and y’. Where is the origin of this new axes, x and y’, with
reference to our normal x and y axes?
In the above we had x  x  2 and y  y  3 which means the origin in this new system is
where x  0 and y  0 . This occurs when x  2  0 and y  3  0 which gives the point
x  2 and y  3 respectively. What does this mean?
Means that the origin of the new system is 2 units to the right and 3 up from our normal
origin.
 x   y  
2 2
b
The hyperbola 2  2  1 gets closer and closer to the two lines y '   x ' where
3 3 a
a  b  3 in our case:
3
y '   x '   x ' which implies y '  x ', y '  x '
3
From above we have x  x  2 and y  y  3 . Substituting these into y '  x ' gives
y '  x '  y  3  x  2  y  x 1
Similarly substituting x  x  2 and y  y  3 into the other line y '   x ' we have

y  x  5
 x   y 
2 2

Hence x 2
 4x  y 2

 6 y  14 is equivalent to  1 which
32 32
is a hyperbola 3 units away from y axis but on the new axis x ' and getting closer and closer
to the lines y   x  5 and y  x  1 (asymptotes) as shown in Fig. 8.

x 2  4 x  y 2  6 y  14
3 3

Figure 8
Chapter 7: Eigenvalues and Eigenvectors 22

G3 Sketching Conics with Cross Product Terms

Why is the topic ‘sketching conics’ placed in this chapter and section?
This is the question that you might be asking yourself since we have not used eigenvalues,
eigenvectors, diagonalization. Up to now we did not need to use the theory of eigenvalues.
Why not?
Because we have dealt with specific quadratics. Do you notice any similarities between the
quadratics sketched so far?
Up to now all the quadratics did not have a cross product term xy which means we did not
need to apply the theory of eigenvalues and eigenvectors. If a quadratic contains a cross
product term such as xy then we are forced into using eigenvalues and eigenvectors. Why?
Because this means the resulting conic has been rotated rather than just translated as the next
example illustrates.
Example 51
Consider the quadratic
10 x 2  8 xy  4 y 2  12
(i) For this quadratic determine the matrix A of the quadratic form xT Ax  12 .
(ii) Find the eigenvalues of A and the orthogonal matrix Q which diagonalizes A.
(iii) Convert the given quadratic into standard form and sketch it.
Solution
(i) The quadratic form matrix A was evaluated in Example 44 of the last section:
 10 4 
10 x 2  8 xy  4 y 2  xT Ax where A   
 4 4
(ii) Similarly the eigenvalues and the orthogonal matrix was found in Example 44:
1 1 2
Eigenvalues 1  2 , 2  12 and orthogonal eigenvector matrix Q   .
5  2 1 
 x'
(iii) Let y    be the new variables of the diagonal form. Again the diagonal form was
 y '
evaluated in Example 44 and we had:
 8 xy  4 y 2  1  x '   2  y ' 
2 2
10 x 2
 2  x '  12  y '  Substituting e.values 1  2 and 2  12
2 2

 8 xy  4 y  2  x '   12  y '   12 . The Right Hand

2 2 2 2
Hence we have 10 x
Side equation 2  x '  12  y '   12 is nearly in standard form for a conic. How do
2 2

convert this into standard form?

Dividing this equation 2  x '  12  y '   12 by 12 gives
2 2

 x '  y '  x '  y '

2 2 2 2

  1 or  1
 
2
6 1 6 12

 x '  y '
2 2

Which conic does this equation   1 represent?

 
2
6 12

 x '  y '
2 2

It is of the format 
 1 which is an ellipse with reference to the new axes x’ and
a2 b2
y’. Where do these axes x’ and y’ lie with respect to our normal x and y axes?
Chapter 7: Eigenvalues and Eigenvectors 23

x  x'
From the last section Proposition (7.38) we had x  Qy where x    , y    and Q was
 y  y '
the above orthogonal matrix (found in Example 44). Remember x’ and y’ were the new
variables or the new axes as we are interested in a sketch.
We can find the new axis x ' by considering any point or vector on this new axis. To find the
directions of the new axes x and y we need to write any points or vectors on these axes in
terms of our normal axes. Generally it is easier to consider the unit vectors 1 0  and
T

 0 1
T
on these axes:

Figure 9
In this new coordinate system the x’ axis is determined by the vector 1 0  which is x '  1
T

and y '  0 . To convert this into our normal xy axes we need to find x  Qy where Q is the
x  x '  1 
above orthogonal matrix, x    our normal axes and y       because x '  1 and
 y  y '  0
y '  0 . Substituting these into x  Qy gives
 x  1 1 2  1   1 1 2 
      Because Q   
 y 5  2 1   0   5  2 1  
1 1
    u  say 
5  2
This means the x’ axes is in the direction of the vector u. Similarly the y’ axes is in the
1 1 2 0
direction given by x  Qy where Q    and y    (the other unit vector,
5  2 1  1 
x '  0 and y '  1 ):
 x  1 1 2 0
    
 y 5  2 1   1 
1  2
    v  say 
5  1 
Our new axes x’ and y’ are in the direction of vectors u and v respectively as shown below in
Fig. 10. What do you notice about vectors u and v?
They are the eigenvectors of matrix A which are the columns of the orthogonal matrix Q.
How does this help in sketching the given quadratic 10 x 2  8 xy  4 y 2  12 ?
 x '  y '
2 2

We have found that this quadratic is equivalent to the ellipse   1 . Thus we

 6
2
12

sketch the ellipse with the centre as x’ and y’ axes and length 6 units along each arm of x’
axis and length 1 unit along each arm of y’ axis as shown below in Fig. 10.
Chapter 7: Eigenvalues and Eigenvectors 24

1
6

Figure 10
Note that the direction of the new axes x’ and y’ is given by the eigenvectors u and v
respectively of matrix A as shown in the above Fig. 10.
Example 51 shows that if we have a cross product term xy then the axes is rotated. How do
we find the new axes x’ and y’?
It is given by the eigenvectors. Why do we need to find eigenvalues and eigenvectors for a
conic with cross-product terms such as xy?
Because the direction of new axes x ' and y ' have changed and are given by the eigenvectors
u and v respectively. Also xT Ax is written in diagonal form 1  x '  2  y ' where 1 and
2 2

2 are the eigenvalues of the matrix A. Remember the diagonal form has no mixed or cross
product terms so we have an equation of a conic.
Next we consider the general case of a conic.
How do we sketch the graph of the general conic
ax 2  2bxy  cy 2  dx  ey  f (*)
where a, b, c, d , e and f are real numbers?
We need to convert this conic (*) first into quadratic form xT Ax and then into diagonal
form. How?
From the last section (F) we know we can place the first three terms of (*), that is
ax 2  2bxy  cy 2 , into quadratic form. How?
x
Let x    then the leading diagonal entries of a matrix A are a and c because these are the
 y
coefficients of x 2 and y 2 respectively. The remaining entries of the matrix are half of the
cross-product term xy which is b (half of 2b). We have
 a b x    a b 
ax 2  2bxy  cy 2   x y     A   
b c  y    b c 
Thus ax 2  2bxy  cy 2  xT Ax (we have assumed b  0 otherwise we will not have a cross-
product term). The remaining terms dx  ey on the Left Hand Side of (*) are not quadratic so
we cannot convert them into quadratic form. How do we place dx  ey into matrix form?
 x
dx  ey   d e   
 y
Let B   d e  . Hence we can write the above general conic in (*) as:
(7.46) ax 2  2bxy  cy 2  dx  ey  xT Ax  Bx  f
 
 x Ax
T  Bx
Chapter 7: Eigenvalues and Eigenvectors 25

By Principle Axes (7.39) of the last section we can write xT Ax in diagonal form as:
xT Ax  y T Dy  1  x '  2  y '
2 2

 x'
where 1 and 2 are the eigenvalues of the matrix A, y    and these x ' , y ' are our new
 y '
axes. Note that converting to diagonal form 1  x '  2  y ' gives an equation of a
2 2

conic which we can sketch.

If we are going to use this diagonal form then we also need to convert Bx into the new axes
x ' and y ' . How?
By Proposition (7.38) of the last section we know that x  Qy where Q is the orthogonal
matrix. Substituting this x  Qy into Bx where B   d e  gives
 x'
Bx  B  Qy    BQ  y   BQ   
 y '
Let BQ   d ' e ' then we have
 x'
Bx  d ' e '    d ' x '  e ' y '
 y '
Hence substituting these xT Ax  1  x '  2  y ' and Bx  d ' x ' e ' y ' into (7.46) gives
2 2

(7.47) ax 2  2bxy  cy 2  dx  ey  1  x '2  2  y '2  d ' x '  e ' y '  f

We use the Right Hand Side of (7.47) to sketch a conic of the form
ax 2  2bxy  cy 2  dx  ey  f
Note that we convert from our normal axes x, y to a new pair of axes x, y .

Example 52
Consider the quadratic
2 x 2  2 xy  2 y 2  2 2 x  4 2 y  8
(i) Convert this quadratic into matrix form xT Ax  Bx  f .
(ii) Find the eigenvalues of A and the eigenvector matrix Q which diagonalizes A.
(iii) Convert the given quadratic into the standard form of a conic and sketch it.
Solution
(i) The first three terms of the given quadratic are 2 x 2  2 xy  2 y 2 . How can we write
this in quadratic form?
Let A be the 2 by 2 (because we only have 2 variables) matrix of quadratic form. The matrix
A has both the leading diagonal entries as 2 because the coefficients of x 2 and y 2 is 2. The
other diagonal entry is half the xy coefficient ( 2 ) which is 1 :
 2 1   x 
2 x 2  2 xy  2 y 2   x y    
 1 2   y 
 2 1 
(ii) Our matrix of quadratic form is A    . The eigenvalues  1 and 2  and the
 1 2 
corresponding normalized eigenvectors  u and v  of this matrix A are given by (verify this)
1  1 1  1
1  1, u    and 2  3, v   
2  1 2  1
Chapter 7: Eigenvalues and Eigenvectors 26

How do we convert this 2 x 2  2 xy  2 y 2 into diagonal form?

By the Parallel Axes Theorem (7.39) of the last section:
2 x 2  2 xy  2 y 2  1  x '   2  y ' 
2 2

  x '  3  y ' Substituting the e.values 1  1 and 2  3

2 2

 xT Ax
Axis x’ is in the direction of the eigenvector u and axis y’ is in the direction of eigenvector v.
What is our orthogonal matrix Q equal to?
1 1 1  1  1 1  1 
Q  u v      Because u    and v   
2 1 1   2  1 2  1
Why do we need to find the orthogonal matrix Q?
Because we need to determine Bx  B  Qy  . What is B equal to?
Since our given quadratic is
2 x 2  2 xy  2 y 2  2 2 x  4 2 y  8
Therefore
B  d 
e   2 2 4 2 
because d is the x coefficient 2 2 and e is the y coefficient 4 2 . Substituting these into
Bx  B  Qy  gives:
1 1 1  x ' 
Bx  B  Qy   2 2   4 2 
 
2 1 1   y ' 
1 1  x '   x '

  2 4        2 6     2 x ' 6 y '
Cancelling 2 's 1 1   y '   y '
We have found all the ingredients of xT Ax  Bx  f which means we can write the given
quadratic in the form given in the above formula (7.46). Substituting xT Ax   x '  3  y '  ,
2 2

Bx  2 x ' 6 y ' and f  8 into

(7.46) ax 2  2bxy  cy 2  dx  ey  xT Ax  Bx  f
gives
 2 xy  2 y 2  2 2x  4 2 y   x '  3  y '  2x '  6 y ' 
2 2
2x2 8
On the Right Hand Side we have written the given quadratic in the new x’ and y’ axes. To
sketch this quadratic in the new variables what do we need to do?
(iii) This  x '  3  y '  2 x '  6 y '  8 is not in a recognised format of a
2 2

conic. How do we place this into a standard form of a conic?

Complete the square on this new equation
 x '  3  y '  2 x '  6 y '  8
2 2

so that it is in standard form. Completing the square on each variable separately gives
 x '  2 x '   x ' 1  1
2 2
[Completing the square on x ' terms]
Consider the y ' terms:
Chapter 7: Eigenvalues and Eigenvectors 27

3  y '   6 y '  3  y '   2 y ' Taking Out 3

2 2
 
 3  y ' 1  1 Completing the square on y ' terms
2
 
 3  y ' 1  3
2

Substituting these  x '  2 x '   x ' 1  1 and 3  y '  6 y '  3  y ' 1  3 into the above
2 2 2 2

equation  x '  3  y '  2 x '  6 y '  8 gives

2 2

 x ' 1  1  3  y ' 1  3  8
2 2

 3  y ' 1  x ' 1  12
2 2

Dividing through by 12 and simplifying yields

 x ' 1 3  y ' 1
2 2

  1
12 12
 x ' 1  y ' 1
2 2
 3 1 1
  1  Because 12  4  22 
 12 
2 2
2

x2 y 2
This is nearly in the format of the equation of an ellipse   1 but not quite. How do
a 2 b2
we convert the above into this format?
By letting another pair of variables x '' (x double dashed) and y '' (y double dashed) be
defined by x "  x ' 1 and y "  y ' 1 . Substituting this into the above we have
 x ' 1  y ' 1  x "  y ''
2 2 2 2

    1
 12   12 
2 2
22 22

 x "  y ''
2 2

This   1 now is in standard form of an ellipse with the centre at the

 12 
2
22

crossroads of the new axes x " and y '' . What is the direction of this new axes?
1  1
Well we know x’ is in the direction of the eigenvector u    and y’ in the direction of
2  1
1  1
the other eigenvector v   .
2  1
Where does this brand new axes x " and y '' lie?
From above we have x "  x ' 1 and y "  y ' 1 which means that the origin in this new
system is where x ''  0 and y ''  0 . This occurs where x '  1 and y '  1 . This new axes
x '' is 1 unit below the x’ axis and y '' is 1 unit to the left of the y’ axis as shown below in
Fig. 11.
 x "  y '
2 2

The graph of   1 is an ellipse with centre at the crossroads of x '' and y ''
 
2
12 22

with length of 12 along each arm of x '' axis and length 2 along each arm of y '' axis.
Chapter 7: Eigenvalues and Eigenvectors 28

12

Figure 11
Note that in the above example we had to convert twice from our normal x, y axes in order to
obtain a conic in the correct format so that we could sketch it. Our conversion was:
x, y x, y x, y

SUMMARY
The standard form of a conic is given by:
(7.45) (a) x 2  y 2  a 2 is a circle.
x2 y 2
(b) 2  2  1 is an ellipse.
a b
(c) y  ax 2 is a parabola.
x2 y 2
(d)   1 is a hyperbola.
a 2 b2
We can write a general conic with cross product terms in matrix form as
(7.46) ax 2  2bxy  cy 2  dx  ey  xT Ax  Bx  f
To sketch a conic of this type we need to rewrite this into one of above formats given in
(7.45).
Chapter 7: Eigenvalues and Eigenvectors 29

Exercise 7(g)
1. Sketch the following conics:
(a) x 2  y 2  1 (b) x 2  y 2  9 (c) x 2  y 2  5 (d) x 2  7 y 2  49
(e) x 2  2 y 2  4 (f) 2 x 2  y 2  4 (g) x 2  y 2  4 (h) 2 x 2  y 2  4

2. Sketch the following conics:

(a) x 2  4 y 2  5 (b) 2 x 2  3 y 2  5 (c) x 2  2 y 2  10 (d) 3 x 2  y 2  10

3. By completing the square or otherwise sketch the following curves:

(a) x 2  y 2  2 x  2 y  7 (b) x 2  y 2  4 x  2 y  11
21
(c) 2 x 2  4 y 2  7 x  2 y  (d) 2 x 2  y 2  5 x  2 y  11
8
5
(e) x 2  2 y 2  6 x  8 y  0 (f) 3 x 2  5 y 2  10 x  2 y 
3

4. Sketch the following conics by first writing them in standard form:

(a) 5 x 2  4 xy  2 y 2  36 (b) 2 x 2  2 xy  2 y 2  9 (c) 5 x 2  24 xy  5 y 2  13
(d) 5 x 2  2 12 xy  y 2  49 (e) 5 x 2  2 3 xy  3 y 2  6

5. Sketch the following conics:

2 103
(a) 2 x 2  2 xy  2 y 2  2 y  (b) 5 x 2  4 xy  2 y 2  2 5 x 
3 3
(c) 3 x  2 xy  3 y  4 2 x  13
2 2
(d) 3 x  8 xy  3 y  6 5 x  2 5 y  1
2 2

106 28 7
(e) 10 x 2  8 xy  4 y 2  x y
5 5 2
Brief Solutions to Exercise 7(g)
1. (a), (b) and (c) is a circle with centre (0, 0) and radii 1, 3 and 5 respectively.
(d) Ellipse with centre (0, 0) and crossing the x axis at 7 and y axis at  7
(e) Ellipse with centre (0, 0) and crossing the x axis at 2 and y axis at  2
(f) Ellipse with centre (0, 0) and crossing the x axis at  2 and y axis at 2
(g) Hyperbola with centre origin and crosses the x-axis at 2 and asymptotes y   x
(h) Hyperbola with centre origin and crosses the x-axis at  2 and asymptotes y   2 x
2. (a) Ellipse with centre origin (0, 0) and crosses the x-axis at  5 and y-axis at  5 / 2
(b) Ellipse with centre origin (0, 0) and crosses the x-axis at  5 / 2 and y-axis at  5 / 3
1
(c) Hyperbola with centre origin, crosses the x-axis at  10 and asy y   x
2
(d) Hyperbola with centre origin and crosses the x-axis at  10 / 3 and asy y   3 x
3. (a) A circle of radius 3 and centre  1,  1
(b) A circle of radius 4 with centre  2, 1
For the remaining solutions where it says ‘ along the x’ or y’ axis’ means only one half of
the axis.
Chapter 7: Eigenvalues and Eigenvectors 30

(c) Ellipse with centre (7/4, 1/4) with 3 / 2 along x ' axis and 3 / 2 along y ' axis
(d) Ellipse with centre (5/4, 1) with 11/ 4 along x ' axis and 11/ 8 along y ' axis
x
(e) Hyperbola with centre (3, 2) with 1 along x ' axis and asymptotic to y   0.121 ,
2
x
y  4.121 .
2
5 1
(f) Hyperbola with centre  ,  and 7 / 15 along x ' axis and asymptotes at
3 5
15 15
y x  1.1 , y   x  1.49 .
5 5
1 1 1  2
4. (a) Ellipse with axes x ' and y ' in the direction of u    and v   
5  2 5  1 
respectively with 6 along x ' axis and 6 along y ' axis.
1  1 1  1
(b) Ellipse with axes x ' and y ' in the direction of u    and v   
2  1 2  1
respectively with 3 along x ' axis and 3 along y ' axis.
1  3 1  2 
(c) Hyperbola with axes x ' and y ' in the direction of u    and v   
13  2  13  3 
1
respectively with 1 along x ' axis. Asymptotes at y  5 x, y   x .
5
1  1  1  3
(d) Hyperbola with axes x ' and y ' in the direction of u    and v   
2 3 2 1 
respectively with 7 along y ' axis. Asymptotes at y  6.1x, y  0.82 x .
1 1  1  3
(e) Ellipse with axes x ' and y ' in the direction of u    and v   
2 3 2 1 
respectively with 3 along x ' axis and 1 along y ' axis.
5.
y
y ''
1 y'

y ''
0.5

-1 -0.5 0.5 1 1.5 2 x

(a) -0.5
1
3 1
-1

x'
-1.5
2 2
2x + 2xy + 2y + 2 y = 2 x ''
3
-2 x'

x ''
Chapter 7: Eigenvalues and Eigenvectors 31

y
2 2
5x + 4xy + 2y + 2 5 x = 103
6 3

4 v
y ''
6
2
6
(b) y'

-4 -2 2 4 x
u
-2

-4

x'
-6
x ''

(c)
y y'
2 3 2
3x + 2xy + 3y – 4 2 x = 13
y ''
2

1
v
2
-2 -1 1 2 3 4 x

-1 u 8

-2

x ''
-3
x'
(d)

(e)
Chapter 7: Eigenvalues and Eigenvectors 32

106 28
y
2 2
10x – 8xy + 4y + x– y=7
5 5 2
x ''
4

x'

u
-6 -4 -2 2 x

v y'
-2

-4

y ''

-6

-8
 Complete Solutions 7(g) 1

Complete Solutions 7(g)

1. (a) We are given the conic x 2  y 2  1 . What shape is this conic?
It is a circle with centre origin (0, 0) and radius 1 because we have x 2  y 2  a 2 with a  1 .
Thus the graph of this is given by
x2  y 2  1

(b) What shape is the conic x 2  y 2  9 ?

Again it is a circle with radius 3 and centre origin (0, 0). We have

(c) How do we sketch the graph of the given conic x 2  y 2  5 ?

 5
2
We have a circle with radius equal to 5 because  5 and our a in the general
equation of a circle x 2  y 2  a 2 is 5 . Hence
y

3
5 2 2
x +y =5
2

-3
– 5 -2 -1 1 2 5 3 x

-1

-2

– 5

-3

(d) We are given the conic x 2  7 y 2  49 . Dividing this equation by 49 gives

x2 7 y 2 x 2 7 y 2 x2 y2
  1 simplifies to 2  2  1  2  1
 
2
49 49 7 7 7 7
 Complete Solutions 7(g) 2

x2 y2
What type of graph is given by this equation   1?
 
2
72 7
This is an ellipse with centre origin and crosses the x-axis at 7 and the y-axis at  7 . We
have
y

3 2 2
7 x + 7y = 49

-6 -4 -2 2 4 6 x

-1

-2

– 7-3

(e) What graph is given by x 2  2 y 2  4 ?

x2 2 y 2 x2 y2
If we divide through by 4  22 we have  4    1 which is an
 
2
4 4 22 2
ellipse with centre origin and crosses the x-axis at 2 and y-axis at  2 .
y

2
2 2
2 x + 2y = 4

-2 -1 1 2 x

-1

– 2

-2

(f) How do we sketch the conic 2 x 2  y 2  4 ?

2x2 y 2 x2 y2
Dividing through by 4  22 gives   1 which implies   1 . What graph
 2
2
4 4 22

is given by this equation?

It is an ellipse with centre origin and crosses the x-axis at  2 and y-axis at 2 .
 Complete Solutions 7(g) 3

(g) What graph is given by the equation x 2  y 2  4 ?

x2 y 2
Dividing this equation by 4  22 we can rewrite it as   1 . Do you recognise this
22 22
equation?
It is hyperbola with centre origin (0, 0) and crosses the x-axis at 2 . Also the hyperbola
2
becomes closer and closer to the lines y   x   x . This means y  x and y   x are
2
asymptotes to the hyperbola:

(h) We need to sketch the graph of 2 x 2  y 2  4 . What type of conic is 2 x 2  y 2  4 ?

If we divide through by 4 we have
2x2 y 2 x2 y 2 x2 y2
 1   1   1
 
2
4 4 2 4 2 22

x2 y2
Do you recognise the graph of this equation   1?
 2
2
22

It is a hyperbola with centre origin (0, 0) and crosses the x-axis at  2 . This hyperbola is
2
asymptotic to the lines y   x   2x :
2
 Complete Solutions 7(g) 4

2. (a) What graph is given by the equation x 2  4 y 2  5 ?

Dividing through by 5 gives
x2 4 y 2 x2 4 y2 x2 y2
 1   1   1
       
2 2 2 2
5 5 5 5 5 5 / 22
2 2
x y
What type of conic is   1?
 
2 2
5  5 / 2
 
5
It is ellipse with centre origin (0, 0) and crosses the x-axis at  5 and y-axis at  :
2
y

1.5
5
2
2 2
x + 4y = 5
1

0.5

-3 -2 -1 1 2 3 x
– 5 5

-0.5

-1

5
–
-1.5
2

(b) We need to sketch the graph of 2 x 2  3 y 2  5 . What type of conic does 2 x 2  3 y 2  5

represent?
Dividing through by 5 gives
2x2 3 y 2
 1
5 5
x2 y 2
It looks like an ellipse but the general equation of an ellipse is given by 2  2  1 .
a b
2 2
2x 3y
How do we write the above equation   1 in this format?
5 5
x2 y2  2 1 3 1 
 1  Because  and 
5/ 2 5/3  5 5/ 2 5 5 / 3 
 Complete Solutions 7(g) 5

   
2 2
We can write 5 / 2  5/ 2 and 5 / 3  5 / 3 . Thus we have
x2 y2
2
 2
1
 5 / 2  5 / 3
   
Do you recognise this equation?
It is an equation of an ellipse with centre origin (0, 0) and crosses the x-axis at
5 5 x2 y2
 and y-axis at  . We have 2 x  3 y  5 is equivalent to
2 2
2
 2
1
2 3  5 / 2  5 / 3
   
and this graph is given by:
y

5
3 1.5 2 2
2x + 3y = 5

0.5

-2 -1 1 2 x
5
5 5
–
2 2
-0.5

5
–
-1 3

-1.5

(c) We are given the equation x  2 y  10 . How do we sketch the graph of this equation
2 2

x 2  2 y 2  10 ?
Dividing through by 10 we have
x2 2 y 2 x2 y2
  1 implies that  1
 
2
10 10 10 5

  x2 y2
2
We can write 5 as 5 . We have the equation   1 . How do we sketch the
 10   5 
2 2

graph of this equation?

It is hyperbola with the origin as the centre and crosses the x-axis at  10 . Also the
5 1
hyperbola is asymptotic to y   x x:
10 2
 Complete Solutions 7(g) 6

x 2  2 y 2  10

(d) How do we sketch the graph of 3 x 2  y 2  10 ?

Divide the equation by 10 so that we have
3x 2 y 2 x2 y2
  1 implies that  1
   10 
2 2
10 10 10 / 3
x2 y2
What conic is given by this equation  1?
   10 
2 2
10 / 3
10
It is an hyperbola with centre origin and crosses the x-axis at  . The asymptotes are
3
10
y x   3x
10 / 3

3. (a) How do we sketch the given quadratic x 2  y 2  2 x  2 y  7 ?

By using the method of completing the square. Completing the square on x and y variables
separately gives
x 2  2 x   x  1  12
2
[Completing the square on x terms]
y 2  2 y   y  1  12
2
[Completing the square on y terms]
 Complete Solutions 7(g) 7

Substituting these x 2  2 x   x  1  12 and y 2  2 y   y  1  12 into the given equation

2 2

x 2  y 2  2 x  2 y  7 yields
 x  1  12   y  1  12  7
2 2

9 x  1   y  1
2 2

Let new variables x’ (x dashed) and y’ (y dashed) be defined by x  x  1 and y  y  1 .

Substituting these into the above we have
 x    y   32
2 2

 x    y    32 ?
2 2
We say x’ and y’ are our new axes. How do we sketch the graph of
 x    y  
 32 is in standard form and fits the equation of the circle of radius 3 but with
2 2

the centre at the crossroads of the axes x and y’. Where is the origin of this new axes, x
and y’, with reference to our normal axes x and y?
In the above we defined the new variables as x  x  1 and y  y  1 which means the origin
in this new system is where x  0 and y  0 . This occurs when x  1  0 and y  1  0
which gives the point x  1 and y  1 respectively. What does this mean?
Means that the origin of the new system x ' y ' is 1 unit left and 1 unit down from our normal
origin. Hence the equation  x    y   32 is a circle of radius 3 and centre  1,  1 :
2 2

(b) We are given the quadratic x 2  y 2  4 x  2 y  11 .

Completing the square on x and y variables separately gives
x 2  4 x   x  2   22
2
[Completing the square on x terms]
y 2  2 y   y  1  12
2
[Completing the square on y terms]
Substituting these x 2  4 x   x  2   22 and y 2  2 y   y  1  12 into the given equation
2 2

x 2  y 2  4 x  2 y  11 yields
 x  2  22   y  1  12  11
2 2

 x  2    y  1
 16
2 2

Let new variables x’ (x dashed) and y’ (y dashed) be defined by x  x  2 and y  y  1 .

Substituting these into the above we have
 x    y   16  42
2 2

 x    y    42 ?
2 2
We say x’ and y’ are our new axes. How do we sketch the graph of
 x    y  
 42 is an equation of a circle of radius 4 and with the centre at the crossroads of
2 2

the axes x and y’. Where is the origin of this new axes, x and y’, with reference to our
normal axes x and y?
 Complete Solutions 7(g) 8

In the above we defined the new variables as x  x  2 and y  y  1 which means the origin
in this new system is where x  0 and y  0 . This occurs when x  2  0 and y  1  0
which gives the point x  2 and y  1 respectively. What does this mean?
Means that the origin of the new system x ' y ' is at the point  2, 1 .
Hence the equation  x    y   42 is a circle of radius 4 with centre  2, 1 :
2 2

y y' 2 2
x + y – 4x – 2y = 11
2 2
x + y of– 4Intersection
Point x – 2y = 11
(2,1) 4

2
x'

-2 2 4 6 x

-2

-4

21
(c) We are given the conic 2 x 2  4 y 2  7 x  2 y 
. How do we sketch this?
8
First we complete the square on each variable separately. Completing the square on the x
variable:
 7 
2x2  7 x  2  x2  x  Taking out 2
 2 
 7 7 
2 2

 2  x       Completing the square

 4   4  

 7  98
2
  7  98 
2

 2 x     Because 2    
 4  16   4  16 
Completing the square on y variable:
 1 
4 y2  2 y  4  y2  y  Taking out 4
 2 
 1 1 
2 2

 4  y       Completing the square

 4   4  

 1 1
2
  1  1
2

 4 y     Because 4    
 4 4  4 4 
 Complete Solutions 7(g) 9

2 2
 7  98  1 1
Substituting these 2 x 2  7 x  2  x    and 4 y 2  2 y  4  y    into the given
 4  16  4 4
21
equation 2 x 2  4 y 2  7 x  2 y  yields
8
2 2
 7  98  1  1 21
2 x     4 y    
 4  16  4 4 8
2 2
 7  1 21 98 1
2 x    4 y       9
 4  4 8 16 4
7 1
Let new variables x’ and y’ be defined by x  x  and y  y  . Substituting these into
4 4
2 2
 7  1
the above 2  x    4  y    9 we have
 4  4
2  x   4  y    9
2 2

How do we sketch the graph of 2  x   4  y   9 ?

2 2

Divide through by 9  32 :
2  x  4  y  
2 2

 1
32 32
 x    y    1
2 2
 2 1 4 1 
 Because 32  32 / 2 and 32  32 / 22 
  3 / 2
2 2
3/ 2

 x   y 
2 2

Do you recognise this equation  1?

  3 / 2
2 2
3/ 2
3
This is the equation of an ellipse with centre x ' and y ' and length along each arm of
2
3
the x ' axis and length along each arm of the y ' axis. What are the coordinates of x ' y '
2
origin?
7 1 7
From above we have x  x  and y  y  therefore the origin of x ' y ' is given by
4 4 4
1 7 1 21
and . Thus the coordinates are  ,  . The given equation 2 x 2  4 y 2  7 x  2 y  is
4 4 4 8
 x   y 
2 2

equivalent to   1 . The graph is given by:

3 / 2  3 / 2
2 2
 Complete Solutions 7(g) 10

y y' x
x '+ y – 5x – 2y = 12
2 2

2 x'

3
1 2
3
2 x'

-1 1 2 3 4 x

 7; 1 
 4 4
 
-1

2 2
2x + 4y – 7x – 2y = 21
-2 8

(d) We are given the quadratic 2 x 2  y 2  5 x  2 y  11 . How do we sketch this?

First we complete the square on each variable separately:
 5 
2 x2  5x  2  x2  x  Taking out 2
 2 
 5 5 
2 2

 2  x       Completing the square

 4   4  
2
 5  50
 2 x   
 4  16
Consider the y terms:
y 2  2 y   y  1  12
2
[Completing the square on y terms]
2
 5  50
Substituting these 2 x  5 x  2  x    and y 2  2 y   y  1  12 into the given
2 2

 4  16
equation 2 x  y  5 x  2 y  11 yields
2 2

2
 5  50
2  x      y  1  12  11
2

 4  16
2
 5 50 242 121
2  x     y  1  11   1  
2

 4 16 16 8
5
Let new variables x’ and y’ be defined by x  x  and y  y  1 . Substituting these into
4
the above we have
121
2  x    y   
2 2

8
121
How do we sketch the graph of 2  x    y  
2 2
?
8
Divide through by 121/8:
2  x   y 
2 2

 1
121/ 8 121/ 8
 x    y    1
2 2

121/16 121/ 8
 Complete Solutions 7(g) 11

 x   y 
2 2

Do you recognise the equation   1?

121/16 121/ 8
 x   y 
2 2

No but we can rewrite this as   1 . Do you recognise this equation?

11/ 4  11/ 8  
2 2

11
This is the equation of an ellipse with centre x ' and y ' and length along each arm of the
4
11
x ' axis and length along each arm of the y ' axis. What are the coordinates of x ' y '
8
origin?
5
From above we have x  x  and y  y  1 therefore the origin of x ' y ' is given by 5 / 4
4
and 1 . Thus the coordinates of the origin of the x ' y ' system are (5/4, 1).
Hence the graph is given by:

11
8

11
4

(e) We are given the quadratic x 2  2 y 2  6 x  8 y  0 . How do we sketch this?

First we complete the square on each variable separately.
x 2  6 x   x  3  32
2
[Completing the square on x terms]
Consider the y terms:
2 y 2  8 y  2  y 2  4 y  Taking out  2
 2  y  2   22  Completing the square on y terms
2
 
 2  y  2   8
2

Substituting these x 2  6 x   x  3  32 and 2 y 2  8 y  2  y  2   8 into the given

2 2

equation x 2  2 y 2  6 x  8 y  0 yields
 x  3  32  2  y  2   8  0
2 2

 x  3
 2  y  2  1
2 2

Let new variables x’ and y’ be defined by x  x  3 and y  y  2 . Substituting these into

the above we have
 Complete Solutions 7(g) 12

 x   2  y   1
2 2

 
 x   y 
2 2

 1  Because 2  1  1 
  
 
2 2
12 1/ 2 1/ 2 1/ 2
 
 x   y 
2 2

How do we sketch the graph of  1?

1/ 2 
2
12

 x '  y '
2 2

This is hyperbola because it is of the format   1 . What are the values of a and b
a2 b2
in this case?
a  1 and b  1/ 2 . What are the coordinates of x ' y ' origin?
From above we have x  x  3 and y  y  2 which means the coordinates of the origin
 x   y 
2 2

are given by (3, 2). The graph of   1 is a hyperbola with centre (3, 2) and
1/ 2 
2
12

length 1 along each arm of the x ' axis. The graph of x 2  2 y 2  6 x  8 y  0 is the same as the
 x   y 
2 2

graph of the hyperbola   1 . We also need to find the asymptotes of this

 
2
12 1/ 2
1/ 2 x'
hyperbola. They are given by y '   x'   . Substituting x  x  3 and y  y  2
1 2
x'
into the first of these equations y '  gives
2
x3 x 3 x
y2  y  2  0.121
2 2 2 2
Similarly we have
x3 x 3 x
y2   y  2  4.121
2 2 2 2
x x
The hyperbola is asymptotic to the lines y   0.121 and y    4.121 . The graph is
2 2
given by:

1 1
 Complete Solutions 7(g) 13

5
(f) How do we sketch the quadratic 3 x 2  5 y 2  10 x  2 y  ?
3
First we complete the square on each variable x and y separately. Completing the square on x
variable we have
 10 
3 x 2  10 x  3  x 2  x  Taking out 3
 3 
 10  2  10  2 
 3  x       Completing the square
 6   6  

 5  100
2
  10 
2
100 100 
 3 x     Because 3   3  
 3 12   6 36 12 
Completing the square on the y variable
 2 
5 y 2  2 y  5  y 2  y  Taking out  5
 5 
 2  2 
2 2

 5   y       Completing the square

 10   10  
2 2
 1 4  1 1
 5  y    5  5  y   
 5 100  5 5
2 2
 5  100  1 1
Substituting these, 3 x 2  10 x  3  x    and 5 y 2  2 y  5  y    into the
 3 12  5 5
5
given equation 3 x 2  5 y 2  10 x  2 y  yields
3
2 2
 5  100  1 1 5
3 x     5 y    
 3 12  5 5 3
2 2
 5  1  5 1 100 49
3 x    5 y      
 3  5  3 5 12 5
5 1
Let new variables x’ and y’ be defined by x  x  and y  y  . Substituting these into
3 5
the above we have
49
3  x '  5  y ' 
2 2

5
3  x ' 5  y '
2 2
 49 
 1  Dividing through by
49 / 5 49 / 5  5 
 x '  y '
2 2

 1
49 /15 49 / 25
 x '  y '
2 2

 1
 15   7 5 
2 2
7

 x '  y '
2 2

What conic is given by this equation   1?

 15   7 5 
2 2
7
 Complete Solutions 7(g) 14

Hyperbola with reference to the axes x ' y ' . Where is the origin of these new axes x ' y ' ?
5 1
Since from above we have x  x  and y  y  therefore the origin of the new axes is
3 5
5 1  5 1
where x  and y  which is the coordinate  ,  . The hyperbola given by
3 5 3 5
 x '  y '
2 2
5 1 7
  1 has its centre at  ,  and has a length of along each arm of
7 15   7 5  3 5
2 2
15
the x ' axis as shown below. We also need to find the asymptotes of the hyperbola. Since we
 x '  y '
2 2

have the hyperbola   1 so the asymptotes are given by:

7 15   7 5 
2 2

75 15
y'   x'   x'
7 15 5
5 1 15
Substituting x  x  and y  y  into y '  x ' gives
3 5 5
1 15  5 15 15 1 15
y  x   y x   x  1.1
5 5  3 5 3 5 5
Similarly we have
1 15  5 15 15 1 15
y  x   y x   x  1.49
5 5  3 5 3 5 5
The graph is given by:

7 7
15 15
15
y x  1.1
5
15
y x  1.49
5

4. (a) We are given the quadratic 5 x 2  4 xy  2 y 2  36 . Note that this quadratic contains the
cross-product term xy so we are forced to use eigenvalues and eigenvectors. First we need to
write the given equation in quadratic form. How?
This means that we need to write this as xT Ax where xT   x y  and A is the matrix of
quadratic form. Since we are given 5 x 2  4 xy  2 y 2  36 therefore the leading diagonal
entries of A are 5 and 2. What are the remaining entries?
Half of 4 which is 2 . Thus we have
 Complete Solutions 7(g) 15

 5 2   x    5 2  
5 x 2  4 xy  2 y 2   x y   A   
 2 2   y    2 2  
What else do we need to find?
Find the eigenvalues and the orthogonal matrix Q which diagonalises A. From the previous
sections you should be familiar with deriving eigenvalues, eigenvectors and the orthogonal
matrix Q. Verify that the eigenvalues  1 and 2  and corresponding eigenvectors (u and v)
of A are given by
1 1 1  2
1  1, u    and 2  6, v   
5  2 5  1 
We need to write the given quadratic in diagonal form. How?
 x'
Let y    be the new variables of diagonal form. We have
 y '
y Dy  1  x '  2  y '
T 2 2

  x '  6  y '  Because e.values 1  1 and 2  6

2 2

Hence we have 5 x 2  4 xy  2 y 2   x '  6  y '   36 . Dividing this equation by 36 gives

2 2

 
 x '6  y '  x '  y '
2 2 2 2

  1 or  1  Because 6  1  1 
 6 
 
2 2
36 36 62 36 6 6 
 
 x '  y '
2 2

Which conic does this equation   1 represent?

 6
2
62

Ellipse with reference to the new axes x’ and y’. What is the direction of these axes x’ and
y’?
Remember from the examples from the main text x ' is in the direction of the eigenvector
1 1 1  2
u   and y ' is in the direction of the other eigenvector v   .
5  2 5  1 
How does this help in sketching the given conic 5 x 2  4 xy  2 y 2  36 ?
 x '  y '
2 2

We have found that this conic is equivalent to the ellipse   1 . Thus we sketch
 
2
62 6
the ellipse with the centre as x’ and y’ axes and length 6 along each arm of x’ axis and length
6 along each arm of y’ axis. Hence the graph is given by:
 Complete Solutions 7(g) 16

(b) We are given the quadratic 2 x 2  2 xy  2 y 2  9 . First we need to write the given equation
in quadratic form. How?
This means that we need to write this as xT Ax where xT   x y  and A is the matrix of
quadratic form. We are given 2 x 2  2 xy  2 y 2  9 therefore the leading diagonal entries of A
are 2 and 2. What are the remaining entries?
Half of the xy coefficient (2) which is 1. Thus we have
 2 1  x    2 1 
2 x 2  2 xy  2 y 2   x y     A   
 1 2  y    1 2 
What else do we need to do?
Find the eigenvalues and the orthogonal matrix Q which diagonalises A. Verify that the
eigenvalues  1 and 2  and corresponding eigenvectors (u and v) are given by
1  1 1  1
1  1, u 
  and 2  3, v   
2  1 2  1
We need to write the given quadratic in diagonal form. How?
 x'
Let y    be the new variables of diagonal form. We have
 y '
y Dy  1  x '  2  y '
T 2 2

  x '  3  y '  Because e.values 1  1 and 2  3

2 2

2 x 2  2 xy  2 y 2   x '  3  y '  9 . Dividing this equation by 9 gives

2 2
Hence we have
 
 x '3  y '  x '  y '
2 2 2 2

  1 or  1  Because 3  1  1 
 3 
 
2 2
9 9 32 9 3 3 
 
 Complete Solutions 7(g) 17

 x '  y '
2 2

Which conic does this equation   1 represent?

 3
2
32

Ellipse with its centre at the crossroads of the new axes x’ and y’. What is the direction of
these new axes x’ and y’?
1  1
x ' is in the direction of the normalised eigenvector u    and y ' is in the direction of
2  1
1  1
the other eigenvector v   .
2  1
How does this help in sketching the given conic 2 x 2  2 xy  2 y 2  9 ?
 x '  y '
2 2

We have found that this conic is equivalent to the ellipse   1 . Thus we sketch
 
2
32 3
the ellipse with the centre as x’ and y’ axes and length 3 along each arm of x’ axis and length
3 along each arm of y’ axis. Hence the graph is given by:

(c) We need to sketch the conic 5 x 2  24 xy  5 y 2  13 . First we need to write the given
equation in quadratic form. How?
This means that we need to write this as xT Ax where xT   x y  and A is the matrix of
quadratic form. Since we are given 5 x 2  24 xy  5 y 2  13 therefore the leading diagonal
entries of A are 5 and 5 . What are the remaining entries?
Half of 24 which is 12. Thus we have
5 12   x   5 12  
5 x 2  24 xy  5 y 2   x y     A   
12 5   y    12 5  
What else do we need to find?
Find the eigenvalues and the orthogonal matrix Q which diagonalises A. Verify that the
eigenvalues and eigenvectors of matrix A are given by
1  3 1  2 
1  13, u    and 2  13, v   
13  2  13  3 
We need to write the given quadratic in diagonal form. How?
 x'
Let y    be the new variables of diagonal form. We have
 y '
 Complete Solutions 7(g) 18

y Dy  1  x '  2  y '

T 2 2

 13  x '  13  y '  Because 1  13 and 2  13

2 2

Hence we have 5 x 2  24 xy  5 y 2  13  x '   13  y '   13 . Dividing this equation by 13 gives

2 2

 x '   y '  1
2 2

Which conic does this equation  x '   y '  1 represent?

2 2

Hyperbola with it’s centre at the crossroads of the new axes x’ and y’. What is the direction
of these new axes x’ and y’?
1  3
x ' is in the direction of the eigenvector u    and y ' is in the direction of the other
13  2 
1  2 
eigenvector v   .
13  3 
How does this help in sketching the given conic 5 x 2  24 xy  5 y 2  13 ?
We have found that this conic is equivalent to the hyperbola  x '   y '  1 . Thus we
2 2

sketch the hyperbola with the centre as x’ and y’ axes and length 1 along each arm of x’ axis.
The asymptotes are given by y '   x ' . How do we find the relationship between the new and
old variables?
1  3 2   x'
By using the orthogonal matrix Q    . We have y  Q x where y    is the
T

13  2 3   y '
 x
new axes, and x    is our original axes. Substituting these into y  QT x gives
 y
1  3 2   x  1  3x  2 y 
T T
 x' 1  3 2  x 
           
 y ' 13  2 3   y  13  2 3   y  13  2 x  3 y 
1 1
Substituting the above equations x '   3x  2 y  and y '   2 x  3 y  into the first
13 13
asymptote y '  x ' gives
1 1
 2 x  3 y    3x  2 y 
13 13
Multiplying by 13 gives
2 x  3 y  3 x  2 y  y  5 x
Putting these equations x '  3 x  2 y and y '  2 x  3 y into the other asymptote y '   x '
gives
1
2 x  3 y    3 x  2 y   y   x
5
The graph is given by:
 Complete Solutions 7(g) 19

1 1

(d) How do we sketch the graph of 5 x 2  2 12 xy  y 2  49 ?

We first need to write the given quadratic in the form xT Ax where xT   x y  and A is the
matrix of quadratic form. Since we are given 5 x  2 12 xy  y  49 therefore the leading
2 2

diagonal entries of A are 5 and 1 because these are the coefficients of x 2 and y 2
respectively. What are the remaining entries?
Half of 2 12 which is 12 . Thus we have
 5 12   x    5 12  
5 x 2  2 12 xy  y 2   x y     A   
 12  y    12 
 1    1 
What else do we need to find?
Find the eigenvalues and the orthogonal matrix Q which diagonalises A. Verify that the
eigenvalues and eigenvectors are given by
1  1  1 3
1  1, u    and 2  7, v   
2 3 2 1 
We need to write the given quadratic in diagonal form. How?
 x'
Let y    be the new variables of diagonal form. We have
 y '
y Dy  1  x '  2  y '
T 2 2

   x '  7  y '  Because 1  1 and 2  7

2 2

Hence we have 5 x 2  2 12 xy  y 2    x '   7  y '   49 . Dividing this equation by 49 gives

2 2

 
 x ' 7  y '  x '  y '
2 2 2 2

  1    1  Because 7  1  1 
  
 
2 2
49 49 72 7 49 7 7 
 
 x '  y '
2 2

Which conic does this equation    1 represent?

 
2
72 7
Hyperbola with it’s centre at the crossroads of the new axes x’ and y’. What is the direction
of these new axes x’ and y’?
 Complete Solutions 7(g) 20

1  1 
x ' is in the direction of the eigenvector u    and y ' is in the direction of the other
2 3
1 3
eigenvector v    .
2 1 
How does this help in sketching the given conic 5 x 2  2 12 xy  y 2  49 ?
 x '  y '
2 2

We have found that this conic is equivalent to the hyperbola    1 . Thus we

 7
2
72

sketch the hyperbola with the centre as x’ and y’ axes and length 7 along each arm of y’
axis. We also need to find the asymptotes. These are given by
7 x'
y'   x'  y'  
7 7
1  1 3  x'  x
We have the orthogonal matrix Q    . Substituting y    and x    into
2  3 1   y '  y
y  QT x gives
T
 x '  1  1 3   x  1  1 3  x  1  x  3y 
            
 y ' 2  3 1   y  2  3 1   y  2  3 x  y 
1
Substituting the above equations x '   x  3 y
2
  and y ' 
1
2
 
3 x  y into the first
x'
equation y '  and multiplying by 2 gives
7
x  3y
3x  y 
7
 21x  7 y   x  3 y   7 3 y   
21  1 x

21  1
y x  6.1x
7 3
x'
Putting the above equations into y '   gives
7
 x  3y 
3 x  y      21x  7 y  x  3 y    
7  3 y  1  21 x 
 7 
1  21
y x  0.82 x
7 3
The graph is given by:

7
 Complete Solutions 7(g) 21

(e) We need to sketch 5 x 2  2 3 xy  3 y 2  6 . Multiply this through by 1 which gives

5 x 2  2 3 xy  3 y 2  6
Let’s examine this equation. Writing this in quadratic form gives
 5  3  x    5  3 
5 x 2  2 3 xy  3 y 2   x y     A   
 3 3   y    3 3  
 
The eigenvalues and normalised eigenvectors of the matrix A are (verify this):
1 1  1  3
1  2, u    and 2  6, v   
2 3 2 1 
How do we write the given quadratic 5 x 2  2 3 xy  3 y 2  6 in diagonal form?
5 x 2  2 3 xy  3 y 2  1  x '  2  y ' 
2 2

 2  x '  6  y '  Because 1  2 and 2  6

2 2

Thus we have 5 x 2  2 3 xy  3 y 2  2  x '   6  y '   6 . Dividing this equation by 6 gives

2 2

2  x ' 6  y '  x '

2 2 2

 1    y '  1
2

 3
2
6 6

 x '
2

  y '  1 ?
2
What conic is represented by this equation
 3
2

 x '  y '
2 2

This is an ellipse because it is of the format  1 with a  3 and b  1


a2 b2
respectively. The centre of the ellipse is at the crossroads of the new axes x’ and y’.
1 1 
x ' is in the direction of the normalised eigenvector u    and y ' is in the direction of
2 3
1  3
the other eigenvector v   .
2 1 
How does this help in sketching the given conic 5 x 2  2 3 xy  3 y 2  6 ?
 x '
2

  y '  1 . Thus we sketch

2
We have found that this conic is equivalent to the ellipse
 3
2

the ellipse with the centre as x’ and y’ axes and length 3 along each arm of x’ axis and
length 1 along each arm of y’ axis. The graph is given by:
 Complete Solutions 7(g) 22

2 y 2
-5x + 2 3 xy – 3y = -6
2 x'

y'
1

v u

-2 -1 1 2 x

3 1

-1

-2

5. In each of these cases we need to write the given conic in the format xT Ax  Bx  f .
2
(a) Note that the given equation 2 x 2  2 xy  2 y 2  2 y  is similar to the equation given in
3
question 4(b). This means for transforming the first 3 terms of the Left Hand Side,
2 x 2  2 xy  2 y 2 , into diagonal form we can use the solution of 4(b). We have
2 x 2  2 xy  2 y 2   x '  3  y '  xT Ax
2 2

because the eignvalues and normalised eigenvectors are

1  1 1  1
1  1, u    and 2  3, v   
2  1 2  1
1  1 1
The orthogonal matrix Q    . What is B in the above x Ax  Bx  f equal to?
T

2  1 1
B  d e  where d and e are the coefficients of x and y in the given conic
2
2 x 2  2 xy  2 y 2  2 y 
3
Thus B  0  
2 because there is no x and the y coefficient is 2 . What is Bx equal to?
 x' 1  1 1
From the main text we have Bx  B  Qy  where y    and Q    . We have
 y ' 2  1 1
Bx  B  Qy 
1  1 1  x ' 

 0 2    
2  1 1  y ' 
 1 1  x '   x'
  0 1       1 1     x ' y '
Cancelling 2 ' s  1 1  y '   y '
We have found all the ingredients of xT Ax  Bx  f which means we can write the given
2
quadratic in this form. Substituting xT Ax   x '  3  y '  , Bx   x ' y ' and f 
2 2
into this
3
xT Ax  Bx  f gives
2
2 x 2  2 xy  2 y 2  2 y   x '   3  y '   x ' y ' 
2 2

3
 Complete Solutions 7(g) 23

We have written the given quadratic in the new x’ and y’ axes. To sketch this quadratic in
the new variables what do we need to do?
2
Complete the square on this new equation  x '  3  y '   x ' y '  so that it is in standard
2 2

3
form of a conic. Completing the square on each variable separately gives
2 2

 x '  x '   x '    

2 1 1
[Completing the square on x ' terms]
 2 2
Consider the y’ variable:
 y '
3  y '   y '  3  y '    Taking Out 3
2 2

 3
 1 1 
2 2

 3  y '      Completing the square on y ' terms

 6   6  

 1 1
2
 1
2
1 1
 3  y '    Because 3    3  
 6  12  6 36 12 
2 2
 1 1  1 1
Substituting these  x '  x '   x '   and 3  y '  y '  3  y '  
2 2
into the above
 2 4  6  12
2
equation  x '  3  y '   x ' y '  gives
2 2

3
2 2
 1 1  1 1 2
 x '    3  y '   
 2 4  6  12 3
2 2
 1  1 2 1 1
 x '   3  y '      1
 2  6 3 4 12
We have
2 2 2 2
 1  1  1  1
 x '   y '   x '   y ' 
 2 

6
 2  
2 6
1
1/ 3  
2 2
1 1 1/ 3
x2 y 2
This is nearly the standard format of the equation of an ellipse   1 . How do we
a 2 b2
convert the above into this format?
1 1
By letting another pair of variables x "  x ' and y "  y ' . Substituting this into the
2 6
above we have
2 2
 1  1
 x '   y '   x "
2
 y ''
2
 2 

6
 2  1
   
2 2
12 1/ 3 1 1/ 3
This is an ellipse with the centre at the crossroads of the axes x " and y '' . What is the
direction of the new axes?
 Complete Solutions 7(g) 24

1  1
Well we know x’ is in the direction of the eigenvector u    and y’ is in the direction
2  1
1  1
of the other eigenvector v   .
2  1
But where does this brand new axes x " and y” lie?
1 1
We have x "  x ' and y "  y ' which means that the origin in this new system is where
2 6
1 1
x ''  0 and y ''  0 . This occurs where x '  and y '   . This means that the ellipse is
2 6
1 1
centred at  ,   in the x ' y ' coordinate system as shown below. The graph of
2 6
 x "  y ''
2 2

  1 is an ellipse with centre at the crossroads of x” and y '' with length of 1

 
2
12 1/ 3
along each arm of x” axis and length 1/ 3 along each arm of y '' axis.
y
y ''
1 y'

y ''
0.5

-1 -0.5 0.5 1 1.5 2 x

-0.5
1
3 1
-1

x'
-1.5
2 2
2x + 2xy + 2y + 2 y = 2 x ''
3
-2 x'

x ''
103
(b) We need to sketch 5 x 2  4 xy  2 y 2  2 5 x  . First we convert this into the matrix
3
form xT Ax  Bx  f . What is the matrix A equal to?
Writing the first three terms of the Left Hand Side 5 x 2  4 xy  2 y 2 into xT Ax gives
 5 2 x    5 2 
5 x 2  4 xy  2 y 2  xT Ax   x
y   A   
 2 2 y    2 2 
What are the eigenvalues and normalised eigenvectors of this matrix A?
Verify the following are the eigenvalues and normalised eigenvectors of the matrix A:
1  1 1  2
1  1, u    and 2  6, v   
5  2  5 1
What is xT Ax in diagonal form?
The diagonal form is given by
 Complete Solutions 7(g) 25

xT Ax  1  x '  2  y '
2 2

  x '  6  y '  Because e.values 1  1 and 2  6

2 2

1  1 2
The orthogonal matrix Q   u v     . What is B in the above x Ax  Bx  f
T

5  2 1 
equal to?
B   d e  where d and e are the coefficients of x and y in the given conic
103
5 x 2  4 xy  2 y 2  2 5 x 
3
 
Thus B  2 5 0 because there is no y and the x coefficient is 2 5 . What is Bx equal
to?
 x' 1  1 2
From the main text we have Bx  B  Qy  where y    and Q    . We have
 y ' 5  2 1 
Bx  B  Qy 
1  1 2 x ' 

 2 5 0    
5  2 1   y ' 
 1 2 x '   x'

  2 0       2 4     2 x ' 4 y '
Cancelling 5 ' s  2 1   y '   y '
We have found all the ingredients of xT Ax  Bx  f which means we can write the given
103
quadratic in this form. Substituting xT Ax   x '  6  y '  , Bx  2 x ' 4 y ' and f 
2 2
into
3
this xT Ax  Bx  f gives
2 103
5 x 2  4 xy  2 y 2  x   x '   6  y '   2 x ' 4 y ' 
2 2

5 3
We have written the given quadratic in the new x’ and y’ axes. To sketch this quadratic in the
new variables what do we need to do?
103
Complete the square on this new equation  x '  6  y '  2 x ' 4 y ' 
2 2
so that it is in
3
diagonal form. Completing the square on each variable separately gives
 x '  2 x '   x ' 1  1
2 2
[Completing the square on x ' terms]
Consider the y ' variable:
 4 
6  y '   4 y '  6  y '   y '
2 2

 6 
 4  4 
2 2

 6  y '     
 12   12  
2 2
 1 16  1 2
 6  y '   6  6  y '  
 3 144  3 3
 Complete Solutions 7(g) 26

2
 1 2
Substituting these  x '  2 x '   x ' 1  1 and 6  y '  4 y '  6  y '   into the above
2 2 2

 3 3
103
equation  x '  6  y '  2 x ' 4 y ' 
2 2
gives
3
2
 1  2 103
 x ' 1  1  6  y '   
2

 3 3 3
2

 x ' 1  6  y '    1   36
2 1 103 2
 3 3 3
2
 1
How do we sketch the graph of  x ' 1  6  y '   36 ?
2

 3
First we divide through by 36  6 :2

2 2
 1  1
6 y ' y '
 x ' 1   3    x ' 1   3   1
2 2

 
2
62 62 62 6
x2 y 2
This is nearly in the format of the equation of an ellipse   1 . How do we convert the
a 2 b2
above into this standard format?
1
By letting another pair of variables x "  x ' 1 and y "  y ' . Substituting this into the
3
above we have
2
 1
y '
 x ' 1   3    x "   y ''  1
2 2 2

   
2 2
62 6 62 6
This is an ellipse with the centre at the crossroads of the axes x " and y '' . What is the
direction of the new axes?
1  1
Well we know x’ is in the direction of the eigenvector u    and y’ in the direction of
5  2 
1  2
the other eigenvector v   .
5 1
But where does this brand new axes x " and y” lie?
1
We have x "  x ' 1 and y "  y ' which means that the origin in this new system is where
3
1
x ''  0 and y ''  0 . This occurs where x '  1 and y '   . The ellipse is centred at
3
 x "   y ''  1
2 2
 1
 1,   in the x ' y ' coordinate system as shown below. The graph of
 3
 
2
62 6
is an ellipse with centre at the crossroads of x” and y '' with length of 6 along each arm of x”
axis and length 6 along each arm of y '' axis.
 Complete Solutions 7(g) 27

y
2 2
5x + 4xy + 2y + 2 5 x = 103
6 3

4 v
y ''
6
2
6
y'

-4 -2 2 4 x
u
-2

-4

x'
-6
x ''

(c) We are given the conic 3 x 2  2 xy  3 y 2  4 2 x  13 . First we convert this into the matrix
form xT Ax  Bx  f . What is the matrix A equal to?
Writing the first three terms of the Left Hand Side 3 x 2  2 xy  3 y 2 into xT Ax gives
3 1 x    3 1 
3 x 2  2 xy  3 y 2  xT Ax   x
y   A   
 1 3  y    1 3
What are the eigenvalues and normalised eigenvectors of this matrix A?
Verify the following are the eigenvalues and normalised eigenvectors of the matrix A:
1  1 1  1
1  2, u    and 2  4, v   
2  1 2  1
What is xT Ax in diagonal form?
The diagonal form is given by
xT Ax  1  x '  2  y '
2 2

 2  x '  4  y '  Because 1  2 and 2  4

2 2

1  1 1
The orthogonal matrix Q   u v     . What is B in the above x Ax  Bx  f
T

2  1 1
equal to?
B   d e  where d and e are the coefficients of x and y in the given conic
3 x 2  2 xy  3 y 2  4 2 x  13
 
Thus B  4 2 0 because there is no y and the x coefficient is 4 2 . What is Bx equal
to?
 x' 1  1 1
From the main text we have Bx  B  Qy  where y    and Q    . We have
 y ' 2  1 1
 Complete Solutions 7(g) 28

Bx  B  Qy 
1  1 1  x ' 

 4 2 0    
2  1 1  y ' 
 1 1  x '   x'
  4 0        4 4     4 x ' 4 y '
Cancelling 2 ' s  1 1  y '   y '
We have found all the ingredients of xT Ax  Bx  f which means we can write the given
quadratic in this form. Substituting xT Ax  2  x '  4  y '  , Bx  4 x ' 4 y ' and f  13
2 2

into this xT Ax  Bx  f gives

3 x 2  2 xy  3 y 2  4 2 x  2  x '   4  y '   4 x ' 4 y '  13
2 2

We have written the given quadratic in the new x’ and y’ variables. To sketch this quadratic
in the new variables what do we need to do?
Complete the square on this new equation 2  x '  4  y '   4 x ' 4 y '  13 so that it is in
2 2

standard form of a conic. Completing the square on each variable separately gives
2  x '   4 x '  2  x '   2 x ' Taking out 2
2 2
 
 2  x ' 1  1  2  x ' 1  2
2 2
 
Completing the square on the y ' variable:
4  y '   4 y '  4  y '   y '
2 2
 
 1 1 
2 2
 1
2

 4  y '       4  y '   1
 2   2    2
2
 1
Substituting these 2  x '  4 x '  2  x ' 1  2 and 4  y '  4 y '  4  y '   1 into the
2 2 2

 2
above equation 2  x '  4  y '   4 x ' 4 y '  13 gives
2 2

2
 1
2  x ' 1  2  4  y '   1  13
2

 2
2
 1
2  x ' 1  4  y '   13  3  16
2

 2
2
 1
How do we sketch the graph of 2  x ' 1  4  y '   16 ?
2

 2
First we divide through by 16 :
2 2
 1  1
4  y '  
 x ' 1   2   1
y '
2  x ' 1
2 2

 
2
1 
 
2
16 16 8 22

x2 y 2
This is nearly in the format of the equation of an ellipse   1 . How do we convert the
a 2 b2
above into this format?
 Complete Solutions 7(g) 29

1
By letting another pair of variables x "  x ' 1 and y "  y ' . Substituting this into the
2
above we have
2
 1
 y ' 
 x ' 1   2    x ''   y ''  1
2 2 2

   
2 2
8 22 8 22

This is an ellipse with the centre at the crossroads of the axes x " and y '' . What is the
direction of the new axes?
1  1
Well we know x’ is in the direction of the normalised eigenvector u    and y’ in the
2  1
1  1
direction of the other normalised eigenvector v   .
2  1
But where does this brand new axes x " and y” lie?
1
From above we have x "  x ' 1 and y "  y ' which means that the origin in this new
2
1
system is where x ''  0 and y ''  0 . This occurs where x '  1 and y '  . The ellipse is
2
 1
centred at 1,  in the x ' y ' coordinate system as shown below.
 2
 x ''  y ''
2 2

The graph of   1 is an ellipse with centre at the crossroads of x” and y '' with
 
2
8 22

length of 8 along each arm of x” axis and length 2 along each arm of y '' axis.
y y'
2 3 2
3x + 2xy + 3y – 4 2 x = 13
y ''
2

1
v
2
-2 -1 1 2 3 4 x

-1 u 8

-2

x ''
-3
x'

(d) We need to sketch the given conic 3 x 2  8 xy  3 y 2  6 5 x  2 5 y  1 . First we convert

this into the matrix form xT Ax  Bx  f . What is the matrix A equal to?
Writing the first three terms of the Left Hand Side 3 x 2  8 xy  3 y 2 into xT Ax gives
 Complete Solutions 7(g) 30

 3 4  x    3 4  
3 x 2  8 xy  3 y 2  xT Ax   x
y   A   
 4 3  y    4 3 
What are the eigenvalues and normalised eigenvectors of this matrix A?
Verify the following are the eigenvalues and normalised eigenvectors of the matrix A:
1 1 1  2
1  5, u    and 2  5, v   
5  2 5  1 
What is xT Ax in diagonal form?
The diagonal form is given by
xT Ax  1  x '  2  y '
2 2

 5  x '  5  y '  Because e.values 1  5 and 2  5

2 2

1 1 2
The orthogonal matrix Q   u v     . What is B in the above x Ax  Bx  f
T

5  2 1 
equal to?
B   d e  where d and e are the coefficients of x and y in the given conic
3 x 2  8 xy  3 y 2  6 5 x  2 5 y  1
 
Thus B  6 5 2 5 because the x coefficient is 6 5 and y coefficient is 2 5 . What is
Bx equal to?
 x' 1 1 2
From the main text we have Bx  B  Qy  where y    and Q    . We have
 y ' 5  2 1 
Bx  B  Qy 
1 1 2 x ' 

 6 5 2 5    
5  2 1   y ' 
1 2 x '   x'

 6 2      10 10     10 x ' 10 y '
Cancelling 5 's  2 1   y '   y '
We have found all the ingredients of xT Ax  Bx  f which means we can write the given
quadratic in this form.
Substituting xT Ax  5  x '  5  y ' , Bx  10 x ' 10 y ' and f  1 into this xT Ax  Bx  f
2 2

gives
3 x 2  8 xy  3 y 2  6 5 x  2 5 y  5  x '   5  y '   10 x ' 10 y '  1
2 2

We have written the given quadratic in the new x’ and y’ axes. To sketch this quadratic in the
new variables what do we need to do?
Complete the square on this new equation 5  x '  5  y '  10 x ' 10 y '  1 so that it is in
2 2

standard form. Completing the square on each variable separately gives

5  x '  10 x '  5  x '   2 x '
2 2
 
 5  x ' 1  1  5  x ' 1  5
2 2
 
Completing the square on the y ' variable:
 Complete Solutions 7(g) 31

5  y '  10 y '  5  y '   2 y '

2 2
 
 5  y ' 1  1  5  y ' 1  5
2 2
 
Substituting these 5  x '  10 x '  5  x ' 1  5 and 5  y '  10 y '  5  y ' 1  5 into the
2 2 2 2

above equation 5  x '  10 x ' 5  y '   10 y '  1 gives

2 2

5  x ' 1  5  5  y ' 1  5  1
2 2

5  x ' 1  5  y ' 1  1
2 2

We can rewrite this as

 x ' 1  y ' 1  x ' 1  y ' 1
2 2 2 2

  1 which is  1
   
2 2
1/ 5 1/ 5 1/ 5 1/ 5
x2 y 2
This is nearly in the format of the equation of an hyperbola   1 . How do we convert
a 2 b2
the above into this format?
By letting another pair of variables x "  x ' 1 and y "  y ' 1 . Substituting this into the above
we have
 x ' 1  y ' 1  x ''  y ''
2 2 2 2

   1
       
2 2 2 2
1/ 5 1/ 5 1/ 5 1/ 5
This is a hyperbola with the centre at the crossroads of the axes x " and y '' . What is the
direction of these new axes?
1 1
Well we know x’ is in the direction of the eigenvector u    and y’ in the direction of
5  2
1  2
the other eigenvector v   .
5  1 
But where does this brand new axes x " and y” lie?
We have x "  x ' 1 and y "  y ' 1 which means that the origin in this new system is where
x ''  0 and y ''  0 . This occurs where x '  1 and y '  1 . The hyperbola is centred at
 1, 1 in the new x ' y ' coordinate system as shown below. The graph of
 x ''  y ''
2 2

  1 is a hyperbola with centre at the crossroads of x '' and y '' at a

   
2 2
1/ 5 1/ 5
distance of 1/ 5 on along each arm of the x '' and y axes.
Asymptotes:
 x ''  y ''
2 2

Since we have   1 so the asymptotes are given by:

   
2 2
1/ 5 1/ 5
y   x
Let us use the first of these lines y  x . Substituting x "  x ' 1 and y "  y ' 1 into y  x
gives y ' 1  x ' 1  y '  x ' 2 .
We need to write y ' and x ' in terms of our normal axes x and y. How?
 Complete Solutions 7(g) 32

 x'  x
We use the orthogonal matrix Q because y  QT x where y    and x    . We have
 y '  y
2 x  1  x  2y 
T
 x' 1 1 2  x  1  1
           
 y ' 5  2 1   y  5  2 1   y  5  2x  y 
We have
1 1
x'   x  2 y  and y '   2 x  y 
5 5
Putting these into y '  x ' 2 gives
1 1 x 2 5
 2x  y    x  2 y   2  2x  y  x  2 y  2 5  y  
5 5 3 3
x 2 5
One of the asymptote to the hyperbola is y   . Similarly we can find the other
3 3
asymptote by using y   x . This time we have the asymptote y  3 x . Hence our graph is
given by:

106 28 7
(e) We need to sketch the given conic 10 x 2  8 xy  4 y 2  x y  . First we convert
5 5 2
this into the matrix form x Ax  Bx  f . What is the matrix A equal to?
T

Writing the first three terms of the Left Hand Side 10 x 2  8 xy  4 y 2 into xT Ax gives
 10 4   x    10 4  
10 x 2  8 xy  4 y 2  xT Ax   x
y   A   
 4 4 y    4 4 
What are the eigenvalues and normalised eigenvectors of this matrix A?
Verify the following are the eigenvalues and normalised eigenvectors of the matrix A:
1 1  1  2
1  2, u    and 2  12, v   
5  2 5  1 
What is xT Ax in diagonal form?
The diagonal form is given by
xT Ax  1  x '  2  y '
2 2

 2  x '  12  y '  Because e.values 1  2 and 2  12

2 2
 Complete Solutions 7(g) 33

1 1 2
The orthogonal matrix Q    . What is B in the above x Ax  Bx  f equal to?
T

5  2 1 
B  d e  where d and e are the coefficients of x and y in the given conic:
106 28 7
10 x 2  8 xy  4 y 2 x y
5 5 2
 106 28  106 28
Thus B     because the x coefficient is and y coefficient is  . What is
 5 5 5 5
Bx equal to?
 x' 1 1 2
From the main text we have Bx  B  Qy  where y    and Q    . We have
 y ' 5  2 1 
Bx  B  Qy 
 106 28  1  1 2 x ' 
     
 5 5  5  2 1   y ' 
 106 28   1 2 x '   x'
       10 48     10 x ' 48 y '
 5 5   2 1   y '   y '
We have found all the ingredients of xT Ax  Bx  f which means we can write the given
7
quadratic in this form. Substituting xT Ax  2  x '  12  y '  , Bx  10 x ' 48 y ' and f 
2 2

2
into this xT Ax  Bx  f gives
7
2  x '  12  y '   10 x ' 48 y ' 
2 2

2
We have written the given quadratic in the new x’ and y’ axes. To sketch this quadratic in the
new variables what do we need to do?
7
Complete the square on this new equation 2  x '  12  y '   10 x ' 48 y '  so that it is in
2 2

2
standard form of a conic. Completing the square on each variable separately gives
2  x '  10 x '  2  x '   5 x '
2 2
 
 5 5 
2 2

 2  x '     
 2   2  
2 2
 5 25  5  25
 2  x '   2  2  x '  
 2 4  2 2
Completing the square on the y ' variable:
12  y '  48 y '  12  y '   4 y '
2 2
 
 12  y ' 2   4   12  y ' 2   48
2 2
 
2
 5  25
Substituting these 2  x '  10 x '  2  x '   and 12  y '  48 y '  12  y ' 2   48 into
2 2 2

 2 2
7
the above equation 2  x '  12  y '   10 x ' 48 y '  gives
2 2

2
 Complete Solutions 7(g) 34

2
 5  25 7
2  x '    12  y ' 2   48 
2

 2 2 2
2
 5 7 25
2  x '   12  y ' 2     48  64
2

 2 2 2
Dividing through by 64 gives
2 2
 5  5
2  x '   x ' 
2  12  y ' 2   y ' 2 
2 2
   1 which is  2
 1
   
2 2
64 64 32 4/ 3
x2 y 2
This is nearly in the format of the equation of an ellipse   1 . How do we convert the
a 2 b2
above into this format?
5
By letting another pair of variables x "  x ' and y "  y ' 2 . Substituting this into the
2
above we have
2
 5
 x '   y ' 2 
2
 x ''
2
 y ''
2
 2
   1
      4 / 3
2 2 2 2
32 4/ 3 32
This is an ellipse with the centre at the crossroads of the axes x " and y '' . What is the
direction of the new axes?
1 1
Well we know x’ is in the direction of the normalised eigenvector u    and y’ in the
5  2
1  2
direction of the other normalised eigenvector v   .
5  1 
But where does this brand new axes x " and y” lie?
5
From above we have x "  x ' and y "  y ' 2 which means that the origin in this new
2
5
system is where x ''  0 and y ''  0 . This occurs where x '   and y '  2 . The ellipse is
2
 5 
centred at   ,  2  in the new x ' y ' coordinate system as shown below. The graph of
 2 
 x ''  y ''
2 2

  1 is an ellipse with centre at the crossroads of x '' and y '' with length of
   
2 2
32 4/ 3
32 on along each arm of the x '' axis and length of 4 / 3 on along each arm of the y ''
axis.
 Complete Solutions 7(g) 35

106 28
y
2 2
10x – 8xy + 4y + x– y=7
5 5 2
x ''
4

x'

u
-6 -4 -2 2 x

v y'
-2

-4
y ''

-6

-8