How to Draw a Hyperbolic Paraboloid
Detailed Guide
John Ganci1
Al Lehnen2
1 Richland
College
Dallas, TX
[email protected]
2 Madison
Area Technical College
Madison, WI
[email protected]
August 24, 2011
Introduction
Drawing graphs of the quadric surfaces is fairly straightforward for
paraboloids, ellipsoids, and hyperboloids. However, drawing the
graph of a hyperbolic paraboloid requires some thought.
We describe a set of steps that make drawing the graph of a
hyperbolic paraboloid a routine task.
Throughout the presentation we assume that we have an equation
in 3 variables in the form
Au + Bv 2 + Cw 2 = 0
where A, B, and C are non-zero constants, A and B have the
same sign (that is, AB > 0), C is of opposite sign (that is,
AC < 0), and u, v , w is some permutation of x, y , z.
Note: All coordinates in what follows are assumed to be
uvw -coordinates.
The steps
If we have an equation that satisfies the conditions described on
the previous slide, then we know that its graph is a hyperbolic
paraboloid. We can draw a rough sketch of the graph by following
these steps:
Write the equation in standard form
Identify the axis
Identify two parabolas
Draw the parabolas
Identify two hyperbolas
Draw the hyperbolas
Connect the hyperbolas
Shade the surface
Write in standard form
First write the equation in standard form, which means in the form
u=
v2 w2
− 2
a2
b
where a and b are positive constants.
Examples.
1 The equation 400z + 25x 2 − 16y 2 = 0 can be rewritten as
z=
2
3
y2 x2
y2 x2
−
= 2 − 2
25 16
5
4
so u = z, v = y , w = x, a = 5, and b = 4.
The equation 400z + 25x 2 + 16y 2 = 0 cannot be written in
standard form because neither of the coefficients of the degree
2 terms has sign opposite the sign of the degree 1 term.
The equation z 2 + x − y 2 = 0 can be rewritten as
x = y 2 − z 2 . Here u = x, v = y , w = z, a = 1 = b.
Identify the axis
The axis of a hyperbolic paraboloid is one of the three coordinate
axes. That is, it is either the x-axis, the y -axis, or the z-axis. Once
the equation has been written in standard form, identifying the
axis amounts to identifying the variable of degree 1. In our case,
the axis is the u-axis since u is the variable of degree 1.
Examples.
1
On the previous slide we wrote the equation
2
2
400z + 25x 2 − 16y 2 = 0 as z = y52 − x42 , so the axis is the
z-axis.
2
The equation x = y 2 − z 2 is already in standard form. The
axis is the x-axis.
Comment. Since there is only one variable of degree 1, the reader
might question why we write the equation in standard form. The
reasons why will become apparent on subsequent slides.
Identify the parabolas (1/3)
A hyperbolic paraboloid is, roughly speaking, a surface that is
made up of hyperbolas whose vertices lie on one of two parabolas.
Our next step identifies the two parabolas.
Each parabola arises from setting one of the degree 2 variables to
zero. One of the parabolas, the upper parabola, opens along the
positive u-axis. The other parabola, the lower parabola, opens
along the negative u-axis.
Equation in standard form: u =
The upper parabola: u =
The lower parabola: u =
v2
a2
−
w2
b2
v2
a2
2
− wb2
Example. For x = y 2 − z 2 the upper parabola is x = y 2 and the
lower parabola is x = −z 2 .
Identify the parabolas (2/3)
Each of the parabolas has its vertex at the origin and is symmetric
with respect to the u-axis.
The upper parabola u =
v2
a2
lies in the uv -plane.
2
The lower parabola u = − wb2 lies in the uw -plane.
Example. For x = y 2 − z 2 the upper parabola x = y 2 lies in the
xy -plane. The lower parabola x = −z 2 lies in the xz-plane.
Identify the parabolas (3/3)
We next want to draw arcs for each parabola, each arc being
symmetric with respect to the u-axis. We do this by choosing a
positive number s and then finding the values of the degree 2
2
variable that satisfy 0 ≤ u = va2 ≤ s 2 for the upper parabola and
2
0 ≥ u = − wb2 ≥ −s 2 for the lower parabola.
The u-coordinates of the endpoints of the arc of the upper
parabola are s 2 . The u-coordinates of the endpoints of the arc of
the lower parabola are −s 2 .
Example. For x = y 2 − z 2 we use s = 2. For the upper parabola
x = y 2 we want 0 ≤ x = y 2 ≤ 22 = 4 so −2 ≤ y ≤ 2. For the
lower parabola x = −z 2 we want 0 ≥ x = −z 2 ≥ −22 = −4 so
−2 ≤ z ≤ 2.
Draw the parabolas (1/2)
Draw the upper parabola: x = y 2
z
y
x
−2 ≤ y ≤ 2
0≤x ≤4
Note the upper bound for x
Draw the parabolas (2/2)
Draw the lower parabola: x = −z 2
z
y
x
Note the lower bound for x
−2 ≤ z ≤ 2
−4 ≤ x ≤ 0
Identify two hyperbolas (1/3)
We next associate a hyperbola with each parabola. The hyperbola
associated with the upper parabola is called the upper hyperbola
and the hyperbola associated with the lower parabola is called the
lower hyperbola.
The upper hyperbola lies in the plane u = s 2 , where s is the value
chosen for the arcs of the parabolas. Its vertices are the two
endpoints of the upper parabola. The upper hyperbola is the
2
2
intersection of the hyperbolic cylinder s 2 = va2 − wb2 and the
v2
w2
plane u = s 2 . Note that 1 = (sa)
2 − (sb)2 is an equivalent equation
for the hyperbolic cylinder.
The lower hyperbola lies in the plane u = −s 2 . Its vertices are the
two endpoints of the lower parabola. The lower hyperbola is the
2
2
intersection of the hyperbolic cylinder −s 2 = va2 − wb2 and the
v2
w2
plane u = −s 2 . Note that 1 = (sb)
2 − (sa)2 is an equivalent
equation for the hyperbolic cylinder.
Identify two hyperbolas (2/3)
Since the hyperbolic cylinder associated with the upper hyperbola
is symmetric with respect to both the uv -plane and the uw -plane,
the upper hyperbola is symmetric with respect to two lines: the
first is the intersection of the plane x = s 2 and the uv -plane; the
second is the intersection of the plane x = s 2 and the uw -plane.
Similarly, the lower hyperbola is symmetric with respect to two
lines: the first is the intersection of the plane x = −s 2 and the
uv -plane; the second is the intersection plane x = −s 2 and the
uw -plane.
We next want to draw arcs for both hyperbolas. Finding
appropriate bounds for the arcs is a little more complicated than it
was for the parabolas. We illustrate what to do here with a specific
example. Since we will only be drawing rough sketches of
hyperbolic paraboloids, having exact values is not necessary. The
optional material in Appendix A contains most of the details. See
Appendix B for references to all the details.
Identify two hyperbolas (3/3)
Example. For the equation x = y 2 − z 2 the two parabolas are
Upper: x = y 2 , −2 ≤ y ≤ 2, 0 ≤ x = y 2 ≤ 22 = 4
Lower: x = −z 2 , −2 ≤ z ≤ 2, 0 ≥ x = −z 2 ≥ −22 = −4
Each hyperbola consists of two arcs: an upper arc and a lower arc.
Upper hyperbola: 4 = y 2 − z 2 or 1 =
y2
4
−
z2
4
√
√
Upper arc: y = √
4 + z 2 , −2 ≤ z ≤ 2, 2 ≤ y√≤ 2 2
Lower arc: y = − 4 + z 2 , −2 ≤ z ≤ 2, −2 2 ≤ y ≤ −2
Lower hyperbola: −4 = y 2 − z 2 or 1 =
z2
4
−
y2
4
p
√
Upper arc: z = p
4 + y 2 , −2 ≤ y ≤ 2, 2 ≤ z ≤ 2 2
√
Lower arc: z = − 4 + y 2 , −2 ≤ y ≤ 2, −2 2 ≤ z ≤ −2
The upper hyperbola lies in the plane x = 4 and the lower
hyperbola lies in the plane x = −4.
Draw the hyperbolas (1/2)
Draw the upper hyperbola: 4 = y 2 − z 2 or 1 =
y2
4
z
y
x
Plane: x = 4
−2√
≤z ≤2
√
−2 2 ≤ y ≤ −2 or 2 ≤ y ≤ 2 2
−
z2
4
Draw the hyperbolas (2/2)
Draw the lower hyperbola: −4 = y 2 − z 2 or 1 =
z2
4
z
y
x
Plane: x = −4
−2√
≤y ≤2
√
−2 2 ≤ z ≤ −2 or 2 ≤ z ≤ 2 2
−
y2
4
Connect the hyperbolas
Now we connect the hyperbolas. We do this by drawing four line
segments.
1
Connect the upper hyperbola, upper ends, to the lower
hyperbola, upper ends.
2
Connect the upper hyperbola, lower ends, to the lower
hyperbola, lower ends.
If the arcs of the two hyperbolas are appropriately matched (see
the document An Interesting Property of Hyperbolic Paraboloids)
then these line segments lie on the surface of the hyperbolic
parabolid. (See Appendix B for the location of the document.)
Connect the hyperbolas (1/2)
The pink line segments connect the upper ends.
z
y
x
Connect the hyperbolas (2/2)
The aqua line segments connect the lower ends.
z
y
x
Shade the surface
Draw additional hyperbolas (shown in blue) along the upper
parabola, each parallel to the fixed upper hyperbola. Do the same
(shown in green) for the lower parabola.
z
y
x
Shaded surface with Winplot graph
z
y
x
Summary
We have described a “recipe” you can follow to draw a rough
sketch of a hyperbolic paraboloid. The mathematical details were
intentionally omitted. The interested reader should consult the
appendices for the details.
Appendix
There are two appendices.
Appendix A contains optional material. The material provides most
of the details that were not shown in the body of the presentation.
Appendix B lists references.
Appendix A: Optional material
Optional material
Optional material (1/22)
The optional material slides provide most of the details that were
hinted at in earlier slides but were intentionally omitted. See
Appendix B if you want to see the complete details.
Since the material on these slides is optional, you will provide the
details by working some exercises!
We assume that we are working with the hyperbolic paraboloid
whose equation is
v2 w2
u= 2 − 2
a
b
Optional material (2/22)
Recall that when we looked at drawing the arcs of the upper and
lower parabolas (see the slide whose header is Identify the
parabolas (3/3)), we fixed a value of s and then stated that we
wished to find the values of the degree 2 variables v and w so that
v2
a2
≤ s2
Upper parabola:
u=
Lower parabola:
u = − wb2 ≥ −s 2
2
Exercise 1. Show that for the upper parabola v must satisfy
−sa ≤ v ≤ sa.
Exercise 2. Find a similar inequality for the lower parabola and
show that your inequality is correct.
Optional material (3/22)
Recall that when we looked at drawing the arcs of the upper and
lower hyperbolas (see the slides whose headers are Identify the
hyperbolas (1/3) through Identify the hyperbolas (3/3)), we
did not give many details. The remaining slides and exercises fill in
most of the details.
The following table summarizes what we found for the hyperbolas.
Hyperbola
Plane
Hyperboloid
Upper hyperbola
u = s2
1=
v2
(sa)2
−
w2
(sb)2
Lower hyperbola
u = −s 2
1=
w2
(sb)2
−
v2
(sa)2
Optional material (4/22)
We first look at the upper hyperbola.
Exercise 3. Find the equations for the upper branch and the lower
branch of the upper hyperbola. (Hint: generalize what appears on
the slide whose header
p is Identify the hyperbolas (3/3).)
(Answer: v = ± ba (sb)2 + w 2 . Verify!)
Exercise 4. Restrict the domains of each of the branches of the
upper hyperbola to the closed interval [ −sb, sb ]. What are the
corresponding closed intervals for the ranges of each of the
branches? (The branches, restricted to the closed intervals, are the
arcs.) (Hint:
same
√ hint as for Exercise 3.) (Answer for upper
branch: sa, sa 2 .)
Exercise 5. Verify that your answers to Exercise 3 and Exercise 4
agree with the values shown for the example on the above-named
slide.
Optional material (5/22)
We next look at the lower hyperbola.
Exercise 6. Find the equations for the upper branch and the lower
branch of the lower hyperbola.
Exercise 7. Restrict the domains of each of the branches of the
lower hyperbola to the closed interval [ −sa, sa ]. What are the
corresponding closed intervals for the ranges of each of the
branches?
Exercise 8. Verify that your answers to Exercise 6 and Exercise 7
agree with the values shown for the example on the slide whose
header is Identify the hyperbolas (3/3).
Optional material (6/22)
Note that we have not yet shown why the value s, as used for the
parabolas and hyperbolas, guarantees that the four line segments
do, indeed, lie on the hyperbolic paraboloid. Nevertheless, graphs
shown in the body of the presentation suggest that they are on the
hyperbolic paraboloid.
Before we show why the line segments do lie on the hyperbolic
paraboloid, we first view the equation of the upper hyperbola and
the equation of the lower hyperbola using parametric equations.
Recall that the hyperbolic cosine and the hyperbolic sine are
defined by the equations
cosh(t) =
e t +e −t
2
sinh(t) =
and satisfy
cosh2 (t) − sinh2 (t) = 1
e t −e −t
2
Optional material (7/22)
First we consider the upper hyperbola. Recall that it is the
intersection of the plane u = s 2 and the hyperboloid
w2
v2
1 = (sa)
2 − (sb)2 . The following table shows both the rectangular
equations and the parametric equations.
Upper branch
Lower branch
Rectangular
u = s 2p
v = ba (sb)2 + w 2
u = s2 p
v = − ba (sb)2 + w 2
Parametric
u(t) = s 2
v (t) = sa cosh(t)
w (t) = sb sinh(t)
u(t) = s 2
v (t) = −sa cosh(t)
w (t) = sb sinh(t)
Exercise 9. Verify that the Parametric column is correct.
Optional material (8/22)
Recall that we restricted the domains of each branch of the upper
hyperbola to the closed interval [ −sb, sb ].
√
√
Let p = ln( 2 + 1). Note that p > 0 since 2 + 1 > 1.
Exercise 10. Verify that each of the following is true.
1
2
cosh(0) = 1 and sinh(0) = 0.
√
cosh(p) = 2 = cosh(−p), sinh(p) = 1, and sinh(−p) = −1.
Exercise 11. Verify that each of the following is true.
1
The endpoints
of the upper√arc of the upper hyperbola are
√
(s 2 , sa 2, −sb) and (s 2 , sa 2, sb). The vertex has
coordinates (s 2 , sa, 0). (See Exercise 4.)
2
The parametric equations for the upper branch, restricted to
−p ≤ t ≤ p, map one-to-one onto the upper arc.
Optional material (9/22)
Exercise 12. Verify that each of the following is true.
1
The endpoints
of the lower arc√of the upper hyperbola are
√
(s 2 , −sa 2, −sb) and (s 2 , −sa 2, sb). The vertex has
coordinates (s 2 , −sa, 0). (Use symmetry.)
2
The parametric equations for the lower branch, restricted to
−p ≤ t ≤ p, map one-to-one onto the lower arc.
The table on the next slide summarizes what we’ve found.
Optional material (10/22)
Summary of results for upper hyperbola.
Upper branch
Rectangular
u = s 2p
v = ba (sb)2 + w 2
Bounds for
upper arc
Lower branch
−sb ≤ w ≤ √
sb
sa ≤ v ≤ sa 2
u = s2 p
v = − ba (sb)2 + w 2
Bounds for
lower arc
−sb√≤ w ≤ sb
−sa 2 ≤ v ≤ −sa
Parametric
u(t) = s 2
v (t) = sa cosh(t)
w (t) = sb sinh(t)
−p ≤ t ≤ p
u(t) = s 2
v (t) = −sa cosh(t)
w (t) = sb sinh(t)
−p ≤ t ≤ p
Optional material (11/22)
Exercise 13. Formulate exercises similar to Exercises 11 and 12 to
describe the lower hyperbola. That is, find the endpoints of the
two hyperbolic arcs. Verify that p, as used in those exercises, is
also used for the parametric equations for the arcs of the lower
hyperbola.
The table on the next slide summarizes the results you should get
when you formulate and solve Exercise 13.
Optional material (12/22)
Summary of results for lower hyperbola.
Rectangular
2
u = −sp
w = ba (sa)2 + v 2
Parametric
u(t) = −s 2
v (t) = sa sinh(t)
w (t) = sb cosh(t)
Bounds for
upper arc
Lower branch
−sa ≤ v ≤ sa√
sb ≤ w ≤ sb 2
u = −s 2p
w = − ba (sa)2 + v 2
−p ≤ t ≤ p
Bounds for
lower arc
−sa √
≤ v ≤ sa
−sb 2 ≤ w ≤ −sb
−p ≤ t ≤ p
Upper branch
u(t) = −s 2
v (t) = sa sinh(t)
w (t) = −sb cosh(t)
Optional material (13/22)
We now have everything we need in order to see that the four line
segments lie on the surface of the hyperboloid.
The article An Interesting Property of Hyperbolic Paraboloids
covers a more general case than we’ve covered here. Two values
of s are used, s1 and s2 , for the planes u = s12 and u = −s22 . Two
values of p, p1 and p2 , are used to restrict the parametric
equations for the arcs of the upper and lower hyperbolas.
The critical fact from the article is that the condition
s2 = s1 e (p1 −p2 )
is both necessary and sufficient for the four line segments to lie
precisely on the surface of the hyperbolic paraboloid.
If we choose s1 = s2 then p1 = p2 . This is precisely what we did!
Optional material (14/22)
We finish Appendix A by verifying that the points on the four lines
do, indeed, lie on the hyperbolic paraboloid. The steps shown here
are a modification of what is shown in the article An Interesting
Property of Hyperbolic Hyperboloids.
Note that we said lines, not line segments, in the preceding
paragraph.
The next slide shows the graph of the hyperbolic paraboloid
2
2
u = va2 − wb2 , drawn according to the earlier optional material.
Look back at the slides Optional material (10/22) and Optional
material (12/22) to see the bounds for all the arcs.
Note that the eight endpoints of the hyperbolic arcs have been
labeled P1 through P8 . Note the labels for the axes are u, v , and
w.
Optional material (15/22)
w
P7
P5
P3
P1
v
u
P8
P4
P6
P2
Optional material (16/22)
The values for the coordinates of P1 and P5 are shown in the
following table.
Point
P1
P2
P3
P4
Coordinates
√
( s 2 , sa 2, sb )
Point
P5
P6
P7
P8
Coordinates
√
( −s 2 , sa, sb 2 )
Exercise 14. Fill in the coordinates for the remaining points.
Optional material (17/22)
Recall that if Q1 and Q2 are two specific points in 3-space, we can
find the parametric equations for the line passing through the two
points. One method is described in the following 3 steps.
1
Let Q = (u, v , w ) be an arbitrary point in 3-space.
2
Q is on the line joining Q1 and Q2 if and only if the vectors
−−→
−−−→
−−→
−−−→
Q1 Q and Q1 Q2 satisfy Q1 Q = t Q1 Q2 for some real number t.
3
Set each of the three coordinates equal and solve for u, v , and
w , yielding the desired parametric equations.
Let L15 be the line passing through P1 and P5 . Let L26 be the line
passing through P2 and P6 . Let L37 be the line passing through P3
and P7 . Let L48 be the line passing through P4 and P8 .
We next find the parametric equations of these four lines.
Optional material (18/22)
The parametric equations for L15 are shown in the following table.
Line
Parametric
L15
u = s 2 (2t − 1)
√
v = sa 1 + t( 2 − 1)
√
√
w = sb 2 − t( 2 − 1)
L26
L37
L48
Optional material (19/22)
Exercise 15. Verify that the parametric equations for L15 are
correct. Find the parametric equations for the remaining lines.
We now will verify that all points on these four lines lie on the
hyperbolic paraboloid.
We first verify that all points on L15 lie on the hyperbolic
paraboloid. If (u, v , w ) is on the line, then there is a real number t
such that
u = s 2 (2t
− 1)√
v = sa √
1 + t( 2√− 1) w = sb 2 − t( 2 − 1)
In order to verify that (u, v , w ) lies on the hyperbolic paraboloid
2
2
we must show that u = va2 − wb2 .
Optional material (20/22)
Exercise 16. Verify that each of the following is true.
√
√
2
1 v 2 = s 2 1 + 2t( 2 − 1) + t 2 ( 2 − 1)2 )
a
√ √
√
2
2 w2 = s 2 2 − 2 2t( 2 − 1) + t 2 ( 2 − 1)2
b
2
2
We next show that va2 − wb2 = u, which shows that the point does
lie on the hyperbolic paraboloid.
Optional material (21/22)
Exercise 17. Verify that the following computation is correct.
v2 w2
− 2
a2
b
i
h
√
√
= s 2 1 + 2t( 2 − 1) + t 2 ( 2 − 1)2 )
h
i
√
√ √
−s 2 2 − 2 2t( 2 − 1) + t 2 ( 2 − 1)2
h
√
√
= s 2 1 + 2t( 2 − 1) + t 2 ( 2 − 1)2
i
√
√ √
−2 + 2 2t( 2 − 1) − t 2 ( 2 − 1)2
n
h√
√ √
io
= s 2 −1 + 2t
2−1 + 2
2−1
h
i
√
√
= s 2 −1 + 2t( 2 − 1)( 2 + 1)
= s 2 [−1 + 2t(2 − 1)]
= s 2 (2t − 1)
= u
Optional material (22/22)
Exercise 17 showed that all points on the line L15 lie on the
hyperbolic paraboloid.
Exercise 18. Verify that all points on the other three lines also lie
on the hyperbolic paraboloid. (Hint: for each of the lines, create
and solve exercises similar to Exercise 16 and Exercise 17.)
Appendix A does describe all the steps needed to justify the body
of the presentation. However, Appendix A lacks certain details and
is short on motivation for some of the steps. Consult Appendix B
for references to all the details.
Appendix B: References
There are two documents that contain additional details about the
process described in this presentation. Both are located on the web
page
http://faculty.matcmadison.edu/alehnen/calculus2/Calculus 2 Home Fall 2011.html
Scroll down to the Selected Notes for Calculus II section.
1
John Ganci and Al Lehnen, How To Draw a Hyperbolic
Paraboloid: Quick Guide, 2011.
2
Al Lehnen, An Interesting Property of Hyperbolic Paraboloids,
2009.