SOLUTION: how can I solve this use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3

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Question 717075: how can I solve this use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3
Found 2 solutions by KMST, Edwin McCravy:
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
The equation of a hyperbola with a horizontal transverse axis can be written in the form

I had to look it up (not good with names), but that is what is called the standard form.
(h,k) is the center
Center, vertices, and foci are on the horizontal line
For the vertices,
--> -->
They are at distance from the center, on line
and is the distance between the vertices.
The segment (and the distance) between the vertices is called the transverse axis.

So far we know
, , and -->

The eccentricity is defined based of the focal distance
(distance from each focus to the center) as

and it turns out that , and are related by
We know so
--> -->
Then, plugging that (along with ) into we get
--> --> -->

Finally, plugging the values given (or very easily found) for , , and , plus the hard earned value for into the standard form we get
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
how can I solve this use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3.
You need to know 5 things about hyperbolas to solve this problem:

1. The equation of a hyperbola with horizontal transverse axis is 


2. The length of the transverse axis is , 

3. The eccentricity is ,

4. The center is 

5. 

center(4,1) 

That tells us that h = 4 and k = 1

horizontal transverse axis is 12

That tells us that 2a = 12
                    a = 6

eccentricity 4/3

That tells us that   

Substituting a = 6,  
Cross multiplying,   
                     






So



becomes:


    


Edwin
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