



An
ellipse is just a stretched out circle.












If we put
a minus sign between the terms of an ellipse 


we get a
stretched out hyperbola. 











The
hyperbola still approaches diagonal asymptotes 


that have
the slopes: 











So for
this problem, the slopes are: 











Since we
don't have anything added or subtracted inside the
parentheses, 


the
hyperbola is centered at the origin (0,0). 


That
means the asymptotes are as well. 





That
makes the asymptote equations Y = ^{1}/2_{ }X and Y
=  ^{1}/2_{ }X. 


So the
graph is: 











When we
worked with ellipses, 


we spent
a lot of time talking about focus points. 


Hyperbolas
have focus points too! 


They are
on the centerline of the hyperbola. 


You find
them using the denominators of the X and Y terms. 


Remember
if a term does not have a denominator showing, 


the
denominator is 1. 





The
equation for the distance to the focus point is: 











So for
our hyperbola: 











The
distance from the center to the focus points is: 











The focus
points are (4.47,0) and (4.47,0) 











There is
a trick you can do with hyperbola focus points: 





1) Choose
any point on either hyperbola line. 


2)
Measure the distance from the point to the further away focus point 


3)
Measure the distance to the closer focus point 


4)
Subtract the smaller distance from the larger distance 


5) No
matter what two points you chose, you always get the same answer! 











The
answer will be different for each hyperbola, 


but any
point in a hyperbola will give you the same answer 


as any
other point on that same hyperbola. 





copyright 2005 Bruce Kirkpatrick 
