Matt Luther - Math 2R03 Tutorial - Dec 1, 2011
9.6 #7, 8, 10, 11, 12
9.7 #4, 8, 11, 12
Problem. (9.6.7) Do a translation to get the conic in standard position. Describe
the conic.
(a) 9x2 + 4y 2 − 36x − 24y + 36 = 0
(As a reminder, since there isn’t an xy cross-term we only need handle the
translation, not rotation).
We handle this by grouping x and y terms and completing the square, then doing
a substitution to remove the translation.
First group the terms
(9x2 − 36x) + (4y 2 − 24y) = −36
Then complete the square in each variable
9(x2 − 4x) + 4(y 2 − 6y)
2
2
9(x − 4x + 4) + 4(y − 6x + 9)
= −36
=
−36 + 9 · 4 + 4 · 9 = 36
Now, factoring each part on the left gives
9(x − 2)2 + 4(y − 3)2
So our equation is
9(x − 2)2 + 4(y − 3)2 = 36
Now, do the substitution x0 = x − 2 and y 0 = y − 3 to get
9x02 + 4y 02 = 36
Divide both sides by 36 to get the standard form
y 02
x02
+
=1
4
9
This is an ellipse intersecting the x0 -axis at (−2, 0), (2, 0) and y 0 -axis at (0, −3), (0, 3).
(b) x2 − 16y 2 + 8x + 128y = 256
Group terms
(x2 − 8x) − 16(y 2 + 8y) = 256
Complete the square
(x2 − 8x + 16) − 16(y 2 + 8y + 16) = 256 + 1 · 16 − 16 · 16 = 16
Factor
(x − 4)2 − 16(y + 4)2 = 16
Substitute x0 = x − 4 and y 0 = y + 4
x02 − 16y 02 = 16
Divide to get standard form
x02
− y 02 = 1
16
This is a hyperbola intersecting the x0 -axis at (±4, 0).
1
2
Problem. (9.6.8) Rotate the axes to remove the xy-term. Describe the conic.
(a) 2x2 − 4xy − y 2 + 8 = 0
To remove the xy term we will express the equation in terms of matrix multiplication and diagonalize the matric that deals with the 2nd degree part.
First, note that the left side of the equation can be written as
2 −2
x
x y
+8=0
−2 −1
y
and call the matrix in the middle A.
We can go through diagonalizing A to find that it has eigenvalues λ1 = 3, λ2 =
−2 and corresponding eigenvectors v1 = (−2, 1), v2 = (1, 2). We can normalize
these to get u1 = √15 (−2, 1) and u2 = √15 (1, 2)
Note that we want to diagonalize A with a matrix P which is not only
orthogonal, but has determinant 1, so we need to make sure we put the
vectors in an order which
! ensures this.
Importantly,
−2
√
5
√1
5
√1
5
√2
5
has determinant −1, so we want to arrange the
columns the other way. Make sure you check this, and also be careful
to keep your eigenvalues matched up correctly if you switch the vectors
around.
!
So, let P =
√1
5
√2
5
−2
√
5
√1
5
, which is an orthogonal matrix with determinant 1.
λ2 0
−2 0
Note that the corresponding diagonal matrix is D =
=
. Mind
0 λ1
0 3
the order.
Now, we will substitute ~x = P ~x0 into our equation above. This is a rotation
change of basis.
2 −2
x
x y
+8=0
−2 −1
y
becomes
(P x~0 )T A(P x~0 ) + 8 = 0
The left side simplifies as follows
T
x~0 P T AP x~0 + 8
=
T
x~0 Dx~0 + 8
=
−2x02 + 3y 02 + 8
So our equation, in the new coordinates, is given by
−2x02 + 3y 02 + 8
−2x02 + 3y 02
=
0
= −8
And dividing by −8 to get it in standard form gives
y 02
x02
−
=1
4
8/3
This is a hyperbola intersecting the x0 -axis at (±2, 0)
Problem. (9.6.10) Translate and rotate. Describe the conic. 3x2 − 8xy − 12y 2 −
30x − 64y = 0
3
We need to do the rotation first to get rid o the xy term. We just can’t complete
the square first since we can’t group the terms. We’ll only do the rotation because
there won’t be time to cover this in tutorial, and the numbers get horrible.
So, first, write it in terms of matrices
~xT A~x + K~x = 0
where
A=
3
−4
−4
−12
and
−64
K = −30
We want to orthogonally diagonalize A with a rotation matrix.
Go through that to compute eigenvalues λ1 = −13, λ2 = 4 and corresponding
eigenvectors v1 = (1, 4) and v2 = (−4, 1). Normalize the eigenvectors to get u1 =
√1 (1, 4) and u2 = √1 (−4, 1)
17
17
So, the rotation matrix that diagonalizes A is
!
P =
√1
17
√4
17
−4
√
17
√1
17
Make sure the determinant is +1, and get the corresponding diagonal matrix
(match up the eigenvalues)
−13 0
D=
0
4
Now, do the substitution ~x = P ~x0 to get
~x0T D~x0 + KP ~x0 = 0
Which simplifies to
−13x02 + 4y 02 + KP ~x0 = 0
We need to compute
KP = −30
−64
√1
17
√4
17
−4
√
17
√1
17
!
=
−286
√
17
√56
17
So, simplified further, our equation is
56
286
−13x02 + 4y 02 − √ x0 + √ y 0 = 0
17
17
Then translate, doing another substitution...
Remark 1. The problems in 9.7 are similar, they just have one more variable, so
we deal with matrices which are 3 × 3 or 1 × 3 (for the K matrices) instead. The
idea is still generally the same.
If there’s cross terms, the shape is rotated; use the substitution ~x = P ~x0 where
P is a rotation matrix (orthogonal matrix with determinant 1) that diagonalizes
the quadratic form matrix.
If there’s both a square and a linear term of the same variable, e.g.: x2 , x, the
shape is translated; complete the square to get it in the form (x−k)2 and substitute
x0 = x − k.
If you forget which equation is what shape, it might help to hold one of the
variables constant. This will reduce the equation to something in one less variable,
4
which you might recognize; this reduced equation will describe the “slices” of the
surface along planes parallel to the remaining variables axes. For example:
2
2
z = y4 − x9
When z = k is constant, we get the intersection with planes parallel to the xy2
y2
plane: 4k
− x9k = 1. These describe hyperbolas opening along the y-axis for positive
values of k and along the x-axis for negative values of k.
When x = k is constant, we get the intersection with planes parallel to the yz2
2
plane: z = y4 − k9 . These describe parabolas opening upward in the direction of
the z + -axis.
When y = k is constant, we get the intersection with planes parallel to the xz2
2
plane: z = k4 − x9 . These describe parabolas opening downward in the direction
of the z − -axis.
So, the surface has hyperbolic and parabolic slices.
You’d still have to be familiar-ish with the surfaces or be able to visualize this
based off the information given, but it can help.