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# Hyperbolas

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My handout for the tutoring center. Latex.
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• Name: Vincent
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Hyperbolas
To start our problem we must know a couple pieces of information about Hyperbolas. First equation
(1) is the standard form for a hyperbola. Notice that Hyperbolas have a negative term in their
standard form.
(x - h)2
(y - k)2
-
= 1
(1)
a2
b2
Given any equation we must be able to determine what it is. Equation (2) is in the standard form
for a polynomial, which isn't very helpful for finding out what kind of conic we have. The first
thing that you should notice is that one of our second order terms, x2 and y2, is negative. Knowing
only this we can guess that our equation (1) is most likely a hyperbola. Lets make our equation
(1) into the standard form for a hyperbola. Lets begin by moving all of our constant terms to the
right keeping our varibles on the left.
4x2 - y2 - 24x - 4y + 16 = 0
(2)
We must seperate our variable terms from our constant terms. The standard pratice is to move the
varibles to the left and the constant terms to the right. Notice that I have also grouped up our x
terms and our y terms together, this will help us in our next step.
4x2 - 24x - y2 - 4y = -16
(3)
Next we must complete the square so as to transform our equation further into standard form of
a hyperbola. I have factored a 4 for out of our x terms and a -1 for our y terms so we have nicer
numbers to work with. This is optional but very helpful so I would recomend to do it. Notice the
two spaces in equation (4), this is where we will put some new terms which will make our perfect
square complete.
4(x2 - 6x + []) - (y2 + 4y + []) = -16
(4)
To properly complete the square we must correctly choose a number which makes both out x part
and our y part a perfect square. To do this we begin by dividing middle term by 2, in this case -6
and 4, and then square them it as shown below.
1 x -6
1 x 4
2
2
-32
22
4(x2 - 6x + 9) - (y2 + 4y + 4) = -16
(5)
By adding 9 to the inside we must keep equality by subtracting 9 from our equation. Next since
our 9 was inside the parentheses we must multiple it by the outside 4 to keep equality, this yeilds
36 as shown below.
4(x - 3)2 - 36 - (y + 2)2 + 4 = -16
(6)
We can also see that we did the same thing with our underlined 4, which we multipled by -1. Next
we will move all of our constanct terms over to the right as we did before.
4(x - 3)2 - (y + 2)2 = -16 - 4 + 36
(7)
Now lets divide the entire equation by 16 as our final step.
4(x - 3)2
(y + 2)2
16
-
=
(8)
16
16
16
1

Now we have our equation in standard form for a hyperbola and can begin to pull out the relevent
information.
(x - 3)2
(y + 2)2
-
= 1
(9)
4
16
Lets recall equation (1): (x-h)2 - (y-k)2 = 1. The center of our hyperbola is give by (h, k) so in our
a2
b2
case (3, -2). Notice that the 2 is negative because our equation is written -h and -k. To properly
graph our equation we must next find our a, b and c. We can see that our a2 in this equation is 4

and therefore our a =
4 = 2. In a every hyperbola our a always falls under our non-negative term.

Our b is found similarly: b2 = 16 and b =
16 = 4. Our c is given by the equation a2 + b2 = c2 so
we just solve for c. Our c value is what we use for finding our foci.
c2 = 22 + 42
c2 = 4 + 16

c =
20

c = 2 5
(10)
Now lets graph our hyperbola. First we plot our center at
our (h, k) for our graph it is (3, -2) as we found earlier. Next
we take our a value and move out 2 from our center in both
directions. Since our a value is under the x we move out in
the x direction. Now we will do the same thing for our b
value moving out 4 both up and down since it's under the y
term. Now we place our c is dependant on what term was
first in our equation. If y was first we would move it out in
the y direction. But since x is first, ie not negative, we move
our c out in both x directions from our center. These our
our foci. Now we can draw a box using the a and b points
as guides and next drawing the diagonals as shown in Figure
1. These diagonals our are asymptotes. Now that we have
drawn our picture we can read off all the information that is Figure 1: Setup for (x-3)2 -(y+2)2 = 1
4
16
usually asked for in this type of problem. Also when drawing
our hyperbola depending on whether a was under x or under y it would open up and down or left
and right respectively. In this case our hyperbola would open left and right in the direction of our
foci. Now in general below is a table for any hyperbola and the information that you need to know.
Center
Transverse Axis
Foci
Vertices
Equation
Aymptotes
(h, k)
Parallel to x-axis
(h c, k)
(h a, k)
(x-h)2 - (y-k)2 = 1, a2 + b2 = c2 y - k = a(x - h)
a2
b2
b
(h, k)
Parallel to y-axis
(h, k c)
(h, k a)
(y-h)2 - (x-k)2 = 1, a2 + b2 = c2 y - k = a(x - h)
a2
b2
b
2

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