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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(20059)  (Show Source):
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