Hyperbola 11 science technology engineering and mathematics

Ppt

1. THE
HYPERBOLA

2. 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.
The hyperbola has
two symmetric parts
called branches.
Each branch has a
vertex and a focus.
The axis that contains
the vertices is called
the transverse axis.
This is the same definition as an ellipse except we
have the difference is always constant instead of the
sum.

3. PARTS OF A HYPERBOLA
center
focifoci
c
o
n
j
u
g
a
t
e

a
x
i
s
v
e
rtic
e
s
v
e
r
t
i
c
e
s
The black dashes lines are asymptotes for the graphs.
transverse axis

4. 222
2
2
2
2
where,1 acb
b
y
a
x

The equation for a hyperbola can be derived by using
the definition and the distance formula. The resulting
equation is:
aa
c
b
b
This looks similar to the ellipse equation but notice the sign
difference.
To graph a hyperbola,
make a rectangle that
measures 2a by 2b as
a sketching aid and
draw the diagonals.
These are the
asymptotes.

5. Find the vertices and foci and graph the hyperbola:
1
49
22

yx
From the center the
ends of the
transverse axis are
"a" each direction.
"a" is the square
root of
this value
The ends of
the transverse
axis are the
vertices and
the axis is 2a
long.
(-3, 0) (3, 0)
a a
From the
center the
ends of the
conjugate axis
are "b" each
direction. "b"
is the square
root of this
value
b
b
Make a
rectangle &
draw diagonals
for the
asymptotes.
To find the foci, they are
"c" away from the center
in each direction. Find "c"
by the equation:
222
bac 
49
2
c 6.313c
 0,13 0,13

6. Let's find the equations of the asymptotes.
1
49
22

yx
What is the
slope of this
line?
They are lines with y intercept of 0.
(-3, 0) (3, 0)
a a
Can you see this would
be b over a or in this
case, up two, over three
or 2/3?
b
b
The second
line has the
same slope
only negative.
 0,13 0,13
hint: rise
over run
bmxy 
xy
3
2

mxy

7. The center of the hyperbola may be transformed
from the origin. The equation would then be:
horizontal
transverse
axis

1
2
2
2
2




b
ky
a
hx 
1
2
2
2
2




b
hx
a
ky
v
e
r
t
i
c
a
l

t
r
a
n
s
v
e
r
s
e

a
x
i
s
The axis is determined by the first term NOT by which
denominator is the largest. If the x term is positive it
will be horizontal, if the y term is the positive term it
will be vertical.
The axis is determined by the first term NOT by which
denominator is the largest. If the x term is positive it
will be horizontal, if the y term is the positive term it
will be vertical.

8. Find the center, foci, vertices and graph the ellipse
116834
22
 xyxycomplete the square on the x terms and
then on the y terms
   __3__411__23__24
22
 xxyy
We grouped the y terms and factored out a 4 and grouped the x
terms and factored out a -3. The y squared term is first because it
is positive.
1 11 1
121314
22
 xy
The right hand side must be a 1 so
divide all terms by 12
12 12 12
3 4
1

1
4
1
3
1
22



 xy
This is now in standard
form and we are ready to
find what we need and
graph (next screen)

9. 
1
4
1
3
1
22



 xy
The center is at (h, k).
In this case (1, -1).
a = square root 3 so the vertices (on the transverse axis)
are square root of 3 each way from the center. Since it is
the y term that is positive, we move square root of 3 each
way in the y direction.
this is b
2
so b = 2
(1, -1)
this is a
2
so
a = square
root of 3
To find foci:
222
bac 
743
2
c 6.27c
 71,1
So foci are:
so vertices are: 31,1
Make the rectangle and
asymptotes to help you graph