1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
All traces are ellipses
a = b = c ⇒ sphere
x2 y2 z 2
+
+ =1
a2 b2 c2
Math 114 – Rimmer
12.6 Quadric Surfaces
Horizontal traces are ellipses
Vertical traces are hyperbolas
Axis of symmetry
corresponds to the
variable with negative
coefficient
x2 y 2 z 2
+
− =1
a 2 b2 c2
1
1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
Vertical traces are hyperbolas
Horizontal traces are ellipses
z = k, k > c
Axis of symmetry
corresponds to the
variable with positive
coefficient
x2 y 2 z 2
− 2 − 2 + 2 =1
a
b
c
Math 114 – Rimmer
12.6 Quadric Surfaces
Horizontal traces are ellipses
Vertical traces are hyperbolas
x = k ⇒ hyperbolas
k ≠0
y = k ⇒ hyperbolas
k = 0 ⇒ lines
Axis of symmetry
corresponds to the
z 2 x2 y 2
=
+
c2 a 2 b2
variable on the left side
2
1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
Horizontal traces are ellipses
Vertical traces are parabolas
Axis of symmetry
corresponds to the
variable raised to
the first power
z x2 y2
=
+
c a2 b2
Math 114 – Rimmer
12.6 Quadric Surfaces
Horizontal traces are hyperbolas
Vertical traces are parabolas
z x2 y2
=
−
c a 2 b2
3
1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
x2 y 2
+
=1
a 2 b2
Math 114 – Rimmer
12.6 Quadric Surfaces
y2 x2
−
=1
a2 b2
4
1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
y = ax 2
Math 114 – Rimmer
12.6 Quadric Surfaces
VII
IV
II
III
VI
I
VIII
V
5
1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
y2
31. x =
1
2
+
z2
1
3
x y2 z2
=
+
6 3
2
or
elliptic paraboloid
vertex: ( 0, 0, 0 )
opens in the positive x − direction
y2 z2
−
4 1
32. x =
hyperbolic paraboloid
Math 114 – Rimmer
12.6 Quadric Surfaces
36 = −36 + ___
4 + 4 z 2 − 24 z + ___
4 + ___
36
33. 4x 2 + y 2 − 4 y + ___
(
) (
)
36
4 + 4 z 2 − 6 z + ___
9 = −36 + ___
4 + ___
4x 2 + y 2 − 4 y + ___
2
2
4x 2 + ( y − 2 ) + 4 ( z − 3) = 4
2
2
x 2 ( y − 2 ) ( z − 3)
+
+
=1
1
4
1
ellipsoid
center:
( 0, 2,3)
6
1/17/2012
Math 114 – Rimmer
12.6 Quadric Surfaces
34. 4 y 2 − 16 y + ___
16 z 2 − 4 z + ___
16 + ___
4 = −20 + x + ___
4
(
) (
)
16 + ___
4 y 2 − 4 y + ___
4 + z 2 − 4 z + ___
4
4 = −20 + x + ___
2
2
4 ( y − 2) + ( z − 2) = x
2
x ( y − 2) ( z − 2)
=
+
4
1
4
2
elliptic paraboloid
vertex:
( 0, 2, 2 )
Math 114 – Rimmer
12.6 Quadric Surfaces
Summary:
x2 y2 z 2
+
+
=1
a 2 b2 c2
Ellipsoid
x2 y2 z2
+
− =1
a 2 b2 c 2
Hyperboloid of one sheet
2
−
2
all variables present
2
x
y
z
− 2 + 2 =1
2
a
b
c
z 2 x2 y2
=
+
c2 a 2 b2
Hyperboloid of two sheets
all variables squared
Cone ( elliptic )
z x2 y 2
=
+
c a 2 b2
Paraboloid ( elliptic )
z x2 y2
=
−
c a2 b2
Paraboloid ( hyperbolic )
all variables present
one variable
not squared
one variable not present ⇒ cylinder opening in the direction of the missing variable
x2 y2
x2 z 2
z = ax 2 Parabolic cylinder
+ 2 = 1 Elliptic cylinder
− = 1 Hyperbolic cylinder
2
a
b
a 2 b2
7