10.5
Lines
and
Planes
Conics
I.
Parabolas
Equation
Axis
Opens
x 2 = 4 py y‐axis
up
x 2 = !4 py y‐axis
down
y 2 = 4 px x‐axis
right
y 2 = !4 px x‐axis
left
2
2
x
=
−4
y
and
y
= 6x .
Below
are
graphs
of
-3
-2
-1
1
2
3
0.6
-0.5
0.4
0.2
-1.0
-0.2
-1.5
0.5
1.0
1.5
2.0
2.5
3.0
-0.4
-2.0
-0.6
II.
Ellipses
x 2 y2
+ 2 =1
2
b
Ellipses
have
the
equation
a
where
the
center
of
the
ellipse
is
the
origin.
The
ellipse
extends
“a”
units
to
the
left
and
right
of
the
origin
and
“b”
units
above
and
below
the
x 2 y2
+
=1
9
origin.
The
example
below
is
of
4
.
3
2
1
-2
-1
1
2
-1
-2
-3
III.
Hyperbolas
x 2 y2
y2 x 2
! 2 =1
! 2 =1
2
2
b
b
Hyperbolas
are
of
the
form
a
or
a
.
The
hyperbola
crosses
the
axis
of
the
positive
variable.
The
y2
x 2 y2
2
! x = 1!and! !
=1
4
9
examples
below
are
of
16
.
10
5
-3- 2
123
4
-5
2
-4
-2
2
4
-2
-10
-4
3
–
space
z
y
x
Ex.
R2:
x
=
2
y
x
Ex.
R3:
x
=
2
z
y
x
Equation
of
a
plane:
Ax
+
By
+
Cz
=
D
Graph
by
finding
the
x,
y,
and
z
intercepts.
Ex.
2x
+
y
–
z
=
4
z
y
x
Ex.
2x
+
y
=
4
z
y
x
Equation
of
a
plane:
We
need
a
point
P(a,b,c)
and
a
normal
vector
!
!
!
!
n = n1i + n2 j + n3 k n1 x + n2 y + n3 z = n1 , n2 , n3 ! a,b, c
Ex.
Find
the
equation
of
the
plane
through
P(4,
0,
6)
normal
to
!
n = !1, 7, 3 .
Equation
of
a
line:
We
need
a
point
P(a,b,c)
and
a
parallel
vector
!
v = v1 , v2 , v3
x = a + v1t
y = b + v2t
z = c + v3t Ex.
Find
the
equation
of
the
line
through
points
P(4,0,6)
and
Q(1,
1,
‐1)
Ex.
Find
the
equation
of
the
plane
through
the
origin
which
contains
the
line
x = 2+t
y = !1
z = !2t Do:
1.
Find
the
equation
of
the
line
through
P(‐4,
0,
5)
perpendicular
to
the
plane
2y
=
5.
2.
Find
the
equation
of
the
plane
through
the
point
P(‐4,
‐4,
5)
and
parallel
to
the
lines
x = 2 + t !!!!!!!!!!!!!!!!!!!!!!!x = 4 ! 3s
y = !3!!!!!!!!!!!!!!!!!!!!!!!!!!!y = s
z = 3t !!!!!!!!!!!!!!!!!!!!!!!!!!!!z = 6 ! 7s