Download Module 7: Conics Lesson 3 Notes Part 1 Hyperbola Hyperbola

Module 7: Conics
Lesson 3 Notes Part 1
Hyperbola
HyperbolaThe hyperbola is the last conic section we will examine in this unit. As we explore
hyperbolas, you will find that a hyperbola has a lot of similarities to the ellipse.
Examples of hyperbolas can be found when discussing spacecrafts, satellites, radio
systems, and lights. We will discuss real life applications of hyperbolas in the last
lesson of Module 7.
Equation of a hyperbola-
There are two general form equations for a hyperbola:
(x - h)2 ( y - k)2
=1
a2
b2
( y - k)2 (x - h)2
=1
b2
a2
In an ellipse, we had a ‘+’ sign in front of both of the fractions. In a hyperbola, one of
our terms is positive and one is negative…but we always have one of each. That’s
one way we recognize what type of conic section an equation represents. Both the
‘x’ and ‘y’ terms are still squared.
You see two basic equations. Everything is the same except for the placement of the
negative sign. In one equation, the negative sign goes with the ‘y’ term, in the other
it goes with the ‘x’ term.
You’ll notice that the equations are set equal to one, just like in the ellipse.
You’ll notice the letters ‘h’ and ‘k’ appearing again – They have the same meaning as
they did for circles (and ellipses in the last lesson). They tell us the center of the
hyperbola.
‘x’ and ‘y’ still represent the coordinates of ordered pairs on the ellipse
‘a’ and ‘b’ still give us information about how the hyperbola is stretched out
Hyperbola Terminology
Let’s graph a hyperbola. As we do, we’ll learn some vocabulary.
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Graph:
x2 y2

1
16 4
On a piece of graph paper, follow along with these steps:
First, we locate the Center of the hyperbola. Again, we find the center in the same
fashion as that of a circle. The center is (h, k). Here, since there aren’t any values for
‘h’ or ‘k’ (notice the ‘x’ and ‘y’ are by themselves), those values must each be 0. So
the center of our hyperbola is at (0, 0). Put an “X” on (0, 0)
We can now find the value of ‘a’ and ‘b’:
a2 = 16. We take the square root of each side and find that a = 4.
b2 = 4. We take the square root of each side and find that b = 2.
The value under the ‘x’ term was 16. So a = 4. So from our center, we move four
units each direction left and right. This is because the x-axis runs left and right. Put
a small dot on (-4, 0) and another dot on (4, 0)
The value under the ‘y’ term was 4. So b = 2. So from our center, we move two units
each direction down and up. This is because the y-axis runs up and down. Put a dot
on (0, -2) and (0, 2)
These dots start to identify something I like to call “The Box”. Those points are the
midpoints of the sides of a rectangle. You can draw vertical lines through (4, 0) and
(-4, 0) which connect with horizontal lines drawn thought the points (0, 2) and (0, 2). Here’s what “The Box” should look like:
(0, 2)
(-4, 0)
(4, 0)
(0, -2)
The next task is to draw the asymptotes of the hyperbola. Asymptotes still have the
same meaning they always have. They are lines which the hyperbola will get close
to touching, but will never reach.
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The asymptotes are the diagonals of the rectangle you just drew. They extend
infinitely in each direction. We need to find the equations for those asymptotes.
With a center at the origin, our job is a little easier:
Find the equations of the asymptotes of our hyperbola.
Let’s start by finding the asymptote with positive slope. ‘A’ in the diagram:
A
(0, 2)
(-4, 0)
(4, 0)
(0, -2)
Recall from earlier in the course that one way to find the equation of a line is by
simply finding the slope and y-intercept. The asymptotes intersect at the center of
the hyperbola. This is fortunate for us because the center is (0, 0). Therefore, our yintercept is zero.
The slope of our line can be found in one of two ways. We can use the slope formula
along with the two endpoints of the diagonal. We can also just count ‘up and over’.
The two points are (-4, -2) and (4, 2)
m
y2  y1 2  ( 2) 4 1

 
x2  x1 4  ( 4) 8 2
So our slope is ½. Our y-intercept is 0.
Therefore, the equation for Asymptote A is:
y
1
x
2
3
B
A
(0, 2)
(-4, 0)
(4, 0)
(0, -2)
When the center of the hyperbola is (0, 0), the other asymptote equation simply has
the opposite slope of the first.
1
x
2
These procedures are basic and can be followed regardless of where the center is.
We simply calculate the slope of the asymptote using two endpoints of the diagonal.
We can then use point-slope form and write an equation. We’ll look at an example
of that soon.
Our equation for B is therefore:
y
Once we draw our asymptotes, we need to ask ourselves this question:
Does the hyperbola open ‘up and down’ or ‘left and right’?
A hyperbola is a lot like two parabolas. Some will open up/down. Some left/right.
The determining factor is where the negative sign is.
Recall from the study of the parabola that when the squared term was ‘x’, the
parabola opened either up or down. When the squared term was ‘y’, the parabola
opened left or right. You can use that to help you remember this:
When the negative sign is in front of the ‘x’ term, the hyperbola will open up and
down.
When the negative sign is in front of the ‘y’ term, the hyperbola will open left and
right.
In our example, the ‘y’ term is negative. Therefore, this hyperbola opens left and
right.
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The transverse axis is the imaginary line that runs from one section of the
hyperbola to the other. If the hyperbola opens up and down, the transverse axis is
vertical. If the hyperbola opens left and right, the transverse axis is horizontal.
To draw the hyperbola, we first need to label the Vertices. A parabola has a vertex,
a hyperbola has two vertices.
The vertices are always on the box. Remember those four original points you
labeled on your graph? (4, 0), (-4, 0), (0, 2) and (0, -2)? Two of those are the vertices.
Which two? That depends on what direction the hyperbola opens. In this example,
we have established that the hyperbola opens left and right. Therefore, the vertices
are on the left and right side of the box. So our vertices are: (4, 0) and (-4, 0)
NOTE: In an ellipse, the vertices depended on which was bigger, ‘a’ or ‘b’. We
checked for the major axis and put the vertices on the end of that. In a hyperbola, it
doesn’t matter which is bigger, ‘a’ or ‘b’. What matters is which term is negative, the
‘x’ or ‘y’.
To draw a hyperbola, you basically draw a parabolic-type curve twice. You have
identified the vertices. Draw the curve so that it gets closer and closer to the
asymptote in each direction but never touches it:
The hyperbola opens left and right. The vertices are on the box and are labeled
above, (-4, 0) and (4, 0). The edges of the hyperbola get closer and closer to the
asymptotes but never touch.
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Now we need to find the Foci. The foci (pronounced with an ‘s’ sound) are two
special points within the ellipse. To find the foci, we need to know ‘c’. What is ‘c’?
Just a number.
For a hyperbola, we add the two values 𝑎2 and 𝑏 2 , so it doesn’t matter in what order
they appear.
To find the foci of a hyperbola, we take:
a 2  b2  c 2
and solve for ‘c’.
When you find the foci for a hyperbola, move out that many spaces toward the
hyperbola…so if the hyperbola is left and right, that’s where we have the foci. The
foci should always end up within each portion of the hyperbola.
Find the foci of our hyperbola:
x2 y2

1
16 4
a 2  b2  c 2
16  4  c 2
20  c 2
 20  c
𝑐 ≈ 4.472
Since we are moving out a specific
number of spaces, the negative c
value can be ignored.
The center of our hyperbola is (0, 0). The hyperbola opens left and right. Therefore,
we move 20 » 4.472 spaces to the left and right of the center. Our foci are at:
(-4.472, 0) and (4.472, 0):
The foci are within the hyperbola, as they always should be.
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What’s the big deal about the foci?
Pick a point on the hyperbola you’ve drawn. Mark it with a star. Measure the
distance from the star to one of the foci. Record that number.
Now measure the distance from the star to the other foci. Record that number.
This time, however, subtract those two numbers.
Repeat the process for any other point on the hyperbola.
**You should find that the difference of the distances from the star to the two foci is
the same for every point on the hyperbola.
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