Question 1150043: Write an equation for the hyperbola that has eccentricity 2, center at (0,0), and vertex at (0,8)
An equation for the hyperbola is...?
Answer by KMST(5398)   (Show Source):
You can put this solution on YOUR website! FACTS ABOUT HYPERBOLAS:
The equation for a hyperbola centered at (h,k) can be written as
 for some pair (p,q) of positive numbers.
If a given equation can be written as above, you would have the values of h, k, p, and q and the sign for the  part.
For  <-->  <-->  , you see that it must be  -->  -->  -->  ,
so the graph will have an upper branch and a lower branch, like this:
On the other hand, for  <-->  , so it must be  <-->  -->  -->  ,
so there is a left branch and a right branch to the graph, like this:
:  .
In either case, the red and green lines are the asymptotes, with slopes  and  .
Textbooks would write the equation of a hyperbola as
either  for a hyperbola with upper and lower branches with vertices at  ,
or  for a hyperbola with upper and lower branches with vertices at  .
When you write the equation you have number values instead of letters,
but your teacher may expect you to use the same letters they use.
COMPARING THOSE FACTS WITH YOUR QUESTION:
The hyperbola in your question has its center at (0,0), and vertex at (0,8), directly above its center by 8 units.
It is just like the hyperbola with equation or shown in the top drawing above.
The center for the hyperbola in the equation above was (h,k), so for the hyperbola in your question, (h,k) is (0,0), with  .
The distance between the center and the vertex above was labeled as  in the drawing, so for the hyperbola in your question,  or  .
All you need to write your equation is the value of  or  .
ABOUT ECCENTRICITY:
So, what do we know about eccentricity?
It is defined for ellipses and hyperbolas as a ratio of distances from the center, distance to a focus divided by distance to a vertex to the focus.
and for both curves the drawings have right triangles with sides labeled , , and  , or  , , and ,where is the hypotenuse and the distance from center to focus.
According to the Pythagorean theorem, for those triangles  or 
In the drawing above, the triangle side labeled c is not the line segment from the center to a focus, but I draw an arc of circle to show that the lengths are the same.
FINDING AND 
With the equation written as  ,  ,
the eccentricity would be  , so  -->  --> 
Substituting the values found got and ,
becomes  -->  ,  and the equation for the hyperbola in the question,
turns into 
The graph is
 . The asymptotes' slopes are  and  .
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