Transcript Hyberbola

Hyberbola
Conic Sections
Hyperbola

The plane can intersect
two nappes of the cone
resulting in a
hyperbola.
Hyperbola - Definition
A hyperbola is the set of all points in a plane such that
the difference in the distances from two points (foci) is
constant.
| d1 – d2 | is a
constant value.
Finding An Equation
Hyperbola
Hyperbola - Definition
What is the constant value for the difference in the
distance from the two foci? Let the two foci be (c, 0)
and (-c, 0). The vertices are (a, 0) and (-a, 0).
| d1 – d2 | is the constant.
If the length of d2 is
subtracted from the left
side of d1, what is the
length which remains?
| d1 – d2 | = 2a
Hyperbola - Equation
Find the equation by setting the difference in the
distance from the two foci equal to 2a.
| d1 – d2 | = 2a
d1  ( x  c)  y
2
2
d 2  ( x  c) 2  y 2
Hyperbola - Equation
Simplify:
( x  c)2  y 2  ( x  c)2  y 2  2a
Remove the absolute value by using + or -.
( x  c) 2  y 2  ( x  c) 2  y 2  2a
Get one square root by itself and square both sides.
( x  c ) 2  y 2  ( x  c ) 2  y 2  2a

( x  c)  y
2
2
 
2
( x  c )  y  2a
2
2

2
( x  c ) 2  y 2  ( x  c ) 2  y 2  4a ( x  c ) 2  y 2  4a 2
Hyperbola - Equation
( x  c ) 2  y 2  ( x  c ) 2  y 2  4a ( x  c ) 2  y 2  4a 2
Subtract y2 and square the binomials.
x 2  2 xc  c 2  x 2  2 xc  c 2  4a ( x  c) 2  y 2  4a 2
Solve for the square root and square both sides.
4 xc  4a 2  4a ( x  c) 2  y 2
xc  a 2   a ( x  c) 2  y 2
 xc  a 
2 2

  a ( x  c)  y
2
2

2
Hyperbola - Equation
 xc  a 
2 2

  a ( x  c)  y
2
2

2
x 2 c 2  2 xca 2  a 4  a 2  ( x  c ) 2  y 2 
Square the binomials and simplify.
x 2 c 2  2 xca 2  a 4  a 2  x 2  2 xc  c 2  y 2 
x2c2  2xca2  a4  a2 x2  2xca2  a2c2  a2 y 2
x 2 c 2  a 4  a 2 x 2  a 2c 2  a 2 y 2
Get x’s and y’s together on one side.
x 2 c 2  a 2 x 2  a 2 y 2  a 2c 2  a 4
Hyperbola - Equation
x 2 c 2  a 2 x 2  a 2 y 2  a 2c 2  a 4
Factor.
x2  c2  a2   a2 y 2  a2  c2  a2 
Divide both sides by a2(c2 – a2)
x2  c2  a2 
2
2
a2  c2  a2 
a y
 2 2
 2 2
2
2
2
2
a c  a  a c  a  a c  a2 
x2
y2
 2
1
2
2
a c  a 
Hyperbola - Equation
x2
y2
 2
1
2
2
a c  a 
Let b2 = c2 – a2
2
2
x
y
 2 1
2
a b
where c2 = a2 + b2
If the graph is shifted over h units and up k
units, the equation of the hyperbola is:
Hyperbola - Equation
 x  h
a
2
2
y k


b
2
2
1
where c2 = a2 + b2
Recognition:
How do you tell a hyperbola from an
ellipse?
Answer:
A hyperbola has a minus (-) between the
terms while an ellipse has a plus (+).
Graph - Example #1
Hyperbola
Hyperbola - Graph
Graph:
 x  1
9
Center:
2
y  2


16
2
1
(-3, -2)
The hyperbola opens in the
“x” direction because “x” is
positive.
Transverse Axis:
y = -2
Hyperbola - Graph
Graph:
 x  1
9
Vertices
2
y  2


16
2
1
(2, -2) (-4, -2)
Construct a rectangle by
moving 4 units up and
down from the vertices.
Construct the diagonals of
the rectangle.
Hyperbola - Graph
Graph:
 x  1
9
2
y  2


16
2
1
Draw the hyperbola
touching the vertices and
approaching the
asymptotes.
Where are the foci?
Hyperbola - Graph
Graph:
 x  1
2
9
y  2


2
16
1
c 2  a 2  b2
c 2  9  16
c 2  25
c  5
The foci are 5 units from the
center on the transverse axis.
Foci: (-6, -2) (4, -2)
Hyperbola - Graph
Graph:
 x  1
9
2
y  2


2
16
1
Find the equation of the
asymptote lines.
Use point-slope form
y – y1 = m(x – x1)
since the center is on both
lines.
Slope =  4
3
y23
Asymptote Equations
4
3
-4
4
x  1
Graph - Example #2
Hyperbola
Hyperbola - Graph
Sketch the graph without a grapher:
10 y 2  5x2  40 y 10x 15  0
Recognition:
How do you determine the type of conic section?
Answer:
The squared terms have opposite signs.
Write the equation in hyperbolic form.
Hyperbola - Graph
Sketch the graph without a grapher:
10 y 2  5x2  40 y 10x 15  0
10 y  40 y  5x 10x  15
2
2
10  y  4 y  ??   5  x  2 x  ??   15
2
2
10  y  4 y  4   5  x  2 x  1  15  40  5
2
2
10  y  2   5  x  1  50
2
2
10  y  2  5  x  1
50


50
50
50
2
2
Hyperbola - Graph
Sketch the graph
without a grapher:
 y  2
2
5
Center:
x  1


10
2
1
(-1, 2)
Transverse Axis Direction:
Up/Down
Equation:
x=-1
Vertices:
Up/Down 5
from the center or
 1, 2  5 
Hyperbola - Graph
Sketch the graph
without a grapher:
 y  2
5
2
x  1


10
2
1
Plot the rectangular points
and draw the asymptotes.
Sketch the hyperbola.
Hyperbola - Graph
Sketch the graph
without a grapher:
 y  2
5
2
x  1


10
2
1
Plot the foci.
c2  a 2  b2
c 2  10  5
c 2  15
c   15
Foci:
 1, 2 
15

Hyperbola - Graph
Sketch the graph
without a grapher:
 y  2
5
2
x  1


10
2
1
Equation of the asymptotes:
5
y2  
 x  1
10
2
y2  
 x  1
2
Finding an Equation
Hyperbola
Hyperbola – Find an Equation
Find the equation of a hyperbola with foci at (2, 6) and
(2, -4). The transverse axis length is 6.
Conic Section
Recogition
Recognizing a Conic Section
Parabola One squared term. Solve for the term which is not
squared. Complete the square on the squared term.
Ellipse Two squared terms. Both terms are the same “sign”.
Circle Two squared terms with the same coefficient.
Hyperbola Two squared terms with opposite “signs”.