HYPERBOLA - Mr.Zuccheroso

Report
HYPERBOLA
PARTS OF A HYPERBOLA
conjugate axis
The dashed lines are asymptotes for the graphs
transverse axis
center
Focus 1
Focus 2
A hyperbola is the collection of points in the plane
whose difference of distances from two fixed points,
called the foci, is a constant.
P (x,y)
|dPF1 – dPF2|= 2a
Curves open sideways: Horizontal Transverse axis
x2 y2
 2 1
2
a
b
The x-axis contains:
- 2 vertices, (-a, 0) and (a, 0)
- 2 foci at (-c, 0), (c, 0)
- has length = 2a
* the conjugate axis (y-axis) has
length = 2b
For every point P on the hyperbola
we have |dPF1 – dPF2|= 2a
F(-c, 0)
F(c, 0)
Curves open up/down:Vertical Transverse axis.
The y-axis contains:
x2 y2
 2  1
2
a b
- 2 vertices, (0, -b) and (0, b)
- 2 foci at (0, -c), (0, c)
F(0, c)
(0, b)
- has length = 2b
* the conjugate axis (x-axis)
has length = 2a
For every point P on the hyperbola
we have |dPF1 – dPF2|= 2b
(0, -b)
F(0, -c)
HYPERBOLA with any center:
( x  h)
( y  k)

1
2
2
a
b
2
2
( x  h)
( y  k)

1
2
2
a
b
2
2
Pythagorean Property
2
2
2
2
x
y


1
2
2
a
b
or
x
y



1
2
2
a
b
2
2
2
where c  a  b
Graphing Hyperbolas
2
2
2
x
y

1
4 16
Step 1: Make a rectangle
measuring 2a by 2b and draw
the diagonals. These are the
asymptotes.
Step 2: Plot the vertices on
the transverse axis and draw
the arches of the hyperbola
2
x
y
 2 1
2
a
b
b
a
b
c
a
Example 1
2
2
Graph the hyperbola 4x – 16y = 64, and find the focal
distance.
2
2
4x – 16y = 64
64
64
64
x2 y 2
That means a = 4 and b = 2

1
16 4
2
(0, 2)
(–4,0)
(4, 0)
(0,-2)
2
2
c =a +b
2
c = 16 + 4
2
c = 20
c = ±√20
Focal distance is 2√20
Example 2:
Write an equation of the hyperbola centered at (0,0)
whose foci are (0, –6) and (0, 6) and whose vertices
are (0, –4) and (0, 4).
(0, 6)
(0, 4)
(0, –4)
(0, –6)
b= 4 and c = 6
2
2
2
c =a +b
2
2
2
6 =a +4
2
36 = 16 + a
2
20 = a
2
2
x
y

 1
20 16
Example 3: Determine the inequality of the
shaded hyperbolic region with vertex (0,12)
and focus (0,15).
b= 12 and c = 15
2
2
2
c =a +b
2
225 = a + 144
2
81 = a
2
2
x
y

 1
81 144
2
2
x
y

 1
81 144
Example 4: Determine the inequality of the
hyperbola representing the shaded region.
Slope of asymptote
is y = bx
a
Slope of asymptote:
-4-0 = 4
-3-0 3
so b = 4
a 3
b=8
b=4
6 3
2
2
x2 y2
x
y

1

1
36 64
36 64
Example 5: Determine the intersection point(s), if
any, of the line 2x – y + 1 = 0 and the hyperbola
2
2
x
y

1
8 13

similar documents