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The Hyperbola
Appolonius of Perga (225 B.C.) discovered that by intersecting a right circular cone all theway through with a plane parallel to the axis, the intersection provides a resulting curve(conic section) in the shape of a hyperbola.
Ellipses are commonly used in navigation, science, astronomy, business, economics, andrelativistic geometry. The reflective properties of the hyperbola are used in the design oftelescope lenses. The Hubble Space Telescope incorporates a Cassegrain design whichincorporates both a hyperbolic and parabolic mirror.
Hyperbolas help define the Spacetime geometry used in Einstein’s Special Theory ofRelativity
A sonic boom shock wave has the shape of a cone, and it intersects the ground in part ofa hyperbola. It hits every point on this curve at the same time, so that people in differentplaces along the curve on the ground hear it at the same time. Because the airplane ismoving forward, the hyperbolic curve moves forward and eventually the boom can beheard by everyone in its path.
Like an ellipse, a hyperbola has two foci, two vertices, and a centre, but its shape iscompletely different.
The Hyperbola in Terms of a Locus of Points
A hyperbola is the set of points such that the magnitude (absolute value of) of thedifferences between the distances from any point on the curve to two fixed points (thefoci), is constant. This constant value is the length of the transverse axis (thedistance between the vertices).
1 2
2
PF PF k
a k
The dashed lines are the asymptotes of the hyperbola. The asymptotes are not part ofthe actual hyperbola, but the are very useful in drawing the hyperbola.
Note: the foci are outside the hyperbola, whereas in an ellipse, the foci are inside theellipse.
In the ellipse,
1 2PF PF k
Anatomy of a Hyperbola
The centre of the hyperbola is the midpoint of the two foci, F’F. The variable crepresents the distance from the centre to either of the foci.
The vertices, V and V1 are on the same axis as the foci. The distance from thecentre (the midpoint of V1V) of the hyperbola to either of the vertices is thevariable a.
The transverse axis is the line segment whose endpoints are V1 and V. Thelength of the transverse axis is 2a
The conjugate axis is the line segment through the centre that is perpendicularto the transverse axis and whose endpoints are b units from the centre. The lengthof the conjugate axis is 2b.
The Asymptotes are the two lines through the centre of the hyperbola that theends of the hyperbola approach as the curve moves away from the centre. Theasymptotes coincide with the diagonals of the 2a by 2b rectangle.
The variables a, b, and c are related by the equation 2 2 2c a b .
Hyperbolas have the property that if a light ray (or radio signal) passes through (or isaimed at) one focus of the hyperbola and reflects of a hyperbolic surface (mirror) at apoint P, then the reflected signal moves along a straight line determined by the the otherfocus and point P.
If you design a mirror where one focus is the centre of a circle for the celestial sphere,then the light rays will be focus toward the one focus but will bounce off the mirror and allhead toward the other focus.
Equation of a Hyperbola with Centre (0,0) and Transverse Axis on x-axis
The equation of a hyperbola with centre (0,0) and transverse axis on the x-axis is
2 2
2 2 1x ya b
Length of transverse axis: 2aLength of conjugate axis: 2bVertices: ,0a and ,0a
Foci: ,0c and ,0c
Asymptotes:b
y xa
andb
y xa
Equation of a Hyperbola with Centre (0,0) and Transverse Axis on y-axis
The equation of a hyperbola with centre (0,0) and transverse axis on the y-axis is
2 2
2 2 1y xa b
or2 2
2 2 1x yb a
Length of transverse axis: 2aLength of conjugate axis: 2bVertices: ,0a and ,0a
Foci: 0, c and 0,c
Asymptotes:a
y xb
anda
y xb
by xa
by x
a
In this format, the termwith the positive signindicates the axis in whichthe curve opens.
Hyperbolas with Centre at (h, k)
Horizontal Hyperbola Vertical Hyperbola
2 2
2 2 1x h x k
a b
2 2
2 2 1y k x h
a b
Foci: ,h c k and ,h c k Foci: ,h k c and ,h k c
Vertices: ,h a k and ,h a k Vertices: ,h k a and ,h k a
Asymptotes: by x h k
a Asymptotes: a
y x h kb
Example
For the hyperbola 2 29 4 36x y , Find the length of the transverse and conjugate axes,the coordinates of the vertices, and the equations of the asymptotes.
Solution:
The standard form of the equation has a 1 on the right side. Therefore, we divide bothsides of the equation by 36
2 2
2 2
9 4 3636 36 36
14 9
x y
x y
This is the equation of the hyperbola with the transverse axis on the x-axis, since the 2xterm is positive.
From the equation, we have
2
2
4 2
9 3
a a
b b
The length of the transverse axis is 2 2 2 4a
The length of the conjugate axis is 2 2 3 6b
The coordinates of the vertices are ,0 2,0a and ,0 2,0a .
The equation of the asymptotes are32
by x x
a and
32
by x x
a
When drawing the hyperbola, we will use the asymptotes to guide us in thecurves.
First, we use the values of a and b to draw a box with width 2a (4 units) and
height 2b (6 units) centred around the origin 0,0 .
Now we will draw two diagonal lines through the corners of the box. These lines
are the asymptotes32
y x and32
y x
Now draw the hyperbola with vertices at 2,0 and 2,0 , staying within the
asymptotes lines.
Example
Determine and sketch the equation of the hyperbola having foci at 1,6 and 1, 2
and with a conjugate axis of length 4 3 .
Solution:
The hyperbola opens up and down since the foci are on a vertical axis.
The centre of the hyperbola is 1 1 6 2, , 1,2
2 2h k
We have 2 8c so c=4
The value of b is 2 3 .
We can determine a , via the equation 2 2 2
2
2
16 12
42
c a b
a
aa
This gives us the equation of the hyperbola
2 2
2 2
2 2
1
1 21
4 12
x h y ka b
x y
Sketch the box centred at 1,2 with width of 2 4 3b and height of 2 4a .
Now we will draw two diagonal lines through the corners of the box. These lines
are the asymptotes 11 2
3y x and 1
1 23
y x
Now draw the hyperbola with vertices at 0 , 0 1,2 2 1,0h k a and
0 , 0 1,2 2 1,4h k a , staying within the asymptotes lines.
Example
Determine the key properties of 2 29 81x y
Solution:
The fist step is to rewrite the equation in standard form by dividing each term by 81
2 2
2 2
9 8181 81 81
19 81
x y
x y
Since the right side term equals 1, this is a horizontal hyperbola with centre (0, 0)
Lengths of transverse and conjugate axis
2 9a , so 3a 2 81b , so 9b
The length of the transverse axis is 2 2 3 6a
The length of the conjugate axis is 2 2 9 18b
Vertices
Since this is a horizontal hyperbola, the vertices are on the x-axis , 3a units on either
side of the centre (0,0): 3,0 , 3,0 .
Foci
We need the value of c.
2 2 2
9 8190
3 10
c a b
c
Since this is a horizontal hyperbola, the foci are on the x-axis , 3 10c units on either
side of the centre (0,0): 3 10,0 , 3 10,0 .
Asymptotes
The horizontal hyperbola has asymptotes with equations9 33
by x x xa
Example
The long range navigation system (LORAN) uses hyperbolas to enable a ship to determineits exact location by radio. (This technique is becoming obsolete by the implementation ofGSP (Global Positioning System, but it still used as backup).
Two LORAN radio transmitters A and B are situated on a straight line separated by 200km. The two stations simultaneously transmit signals at regular intervals. Determine anequation of a hyperbola described by the flight path of an aircraft when the on board radioreceiver determines that it is 20 km closer to station B than to station A.
Solution:
The property that the difference of the distances from two points (foci) to another point isa fixed value k, is the locus definition of a hyperbola.
Therefore the aircraft , P, is on a hyperbola with foci at F1=(-100, 0) and F2=(100, 0),and the k value is 20.
100 a 100 a
Then the one vertex point ,0a , will have the following property:
distance from first focus to a distance from other focus to a
100 100
20
2 20
10
k
a a k
a a
a
a
Now,
2 2 2
22 2
2
2
100 10
10000 100
10000 1009900
c a b
b
b
b
The equation of the hyperbola is therefore2 2
1100 9900x y