Transcript Sullivan College Algebra Section 7.4

Sullivan Algebra and
Trigonometry: Section 11.4
Objectives of this Section
• Find the Equation of a Hyperbola
• Graph Hyperbolas
• Discuss the Equation of a Hyperbola
• Find the Asymptotes of a Hyperbola
• Work with Hyperbolas with Center at (h,k)
A hyperbola is the collection of all points in
the plane the difference of whose distances
from two fixed points, called the foci, is a
constant.
F1: (-c,0)
F2: (c,0)
Transverse Axis
Equation of a Hyperbola: Center at (0,0);
Foci at (c, 0) and (-c,0)
2
2
x
y


1
2
2
a
b
where b2 = c2 - a2
The transverse axis is the x - axis. The
vertices are at (-a, 0) and (a, 0)
Find the equation of a hyperbola with center
at the origin, one focus at (-5, 0), and a vertex
at (4,0). Graph the equation.
Since the given focus and vertex are on the x-axis,
the transverse axis is the x-axis. The distance from
the center to one of the foci is c = 5.
The distance from the center to one of the vertices
is a = 4. Use c and a to solve for b.
b2 = c2 - a2
b2 = 52 - 42 = 25 - 16 = 9
So, the equation of the hyperbola is:
2
2
x
y

1
16
9
(-4,0)
F1: (-5,0)
(4,0)
F2: (5,0)
2
2
9
7
Discuss the equation: x  y  1
Since the equation is written in the desirable
form, a2 = 9 and b2 = 7
Since b2 = c2 - a2, it follows that c2 = a2 + b2 or
c2 = 9 + 7 = 16. So, the foci are at (4,0) and
(-4,0)
The vertices are at (-3, 0) and (3, 0)
y2
2

x
1
Discuss the equation:
4
Since the x term is subtracted from the y term, the
equation is that of a hyperbola with center at the
origin and transverse axis along the y-axis
The vertices are at (0,2) and (0,-2).
Since b2 = c2 - a2, it follows that c2 = a2 + b2 or
c2 = 4 + 1 = 5. So, the foci are at 0, 5 and
0, 5




2
2
x
y
The hyperbola 2  2  1 has the
a
b
two oblique asymptotes:
b
y x
a
b
y x
a
and
Discuss the equation of the hyperbola
16 x  4 y  16
2
2
Begin by dividing both sides of the equation
by 16 to put the equation in the proper form.
2
2
x y
 1
1 4
The center of the hyperbola is the origin. Since the x
term comes first, the transverse axis is the x-axis.
The vertices are at (1,0) and (-1,0).
Since b2 = c2 - a2, it follows that c2 = a2 + b2 or
c2 = 1 + 4 = 5. So, the foci are at 5 ,0 and
 5 ,0


 
The asymptotes have the equation:
b
y  x  2x
a
and
b
y   x  2 x
a
To graph the hyperbola, form the rectangle
containing the points (a,0), (-a,0), (0,b),
and (0,-b). The extensions of the diagonals
of this rectangle are the asymptotes.
(0,2)

5 ,0

 5 ,0
(-1,0)
(1,0)
(0,-2)
Now, graph the hyperbola using these guides
If an ellipse with center at the origin and
major axis coinciding with a coordinate axis
is shifted horizontally h units and vertically k
units, the resulting ellipse is centered at (h,k)
and has the equation:
( x  h) ( y  k )

1
2
2
a
b
Horizontal
Transverse Axis
( y  k ) ( x  h)


1
2
2
a
b
Vertical
Transverse Axis
2
2
2
2
Find the equation of a hyperbola with
center at (2, -3), one focus at (5, -3), and
one vertex at (3, -3).
The center is at (h,k) = (2, -3). So h = 2 and k = -3
The center, focus, and vertex all lie on the line
y = -3, so the major axis is parallel to the x-axis
and the hyperbola has a horizontal transverse axis
and will have an equation in the form:
( x  h) ( y  k )


1
2
2
a
b
2
2
The distance from the center to the vertex is
a = 1. The distance from the center to the focus
is c = 3. To solve for b,
b2 = c2 - a2
b2 = 32 - 12 = 9 - 1 = 8
So, the equation of the ellipse is:
( x  2) ( y  3)

1
1
8
2
2
The equations of the asymptotes can be found
by shifting the equations for the asymptotes h
units in the horizontal direction and k units in
the vertical direction yielding:
b
y  k  ( x  h)
a
and
b
y  k   ( x  h)
a
For our example:
y  3  2 2 ( x  2)