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A hyperbola centered at the origin (0, 0) crosses the x-axis at (11, 0) and (-11, 0). Which of these could be the equation for this hyperbola?

 Dec 14, 2014

Best Answer 

 #2
avatar+130511 
+10

There are infinite possibilites for this....

The equation will be in the form

x^2 / 121  - y^2 / b^2   = 1        where "a" = 11 and a^2  = 121

Here's a graph when b = 5.........https://www.desmos.com/calculator/ptqorqve0j

Here's a graph when b = 13.........https://www.desmos.com/calculator/tmv8vqrcl6

As "a" stays constant and "b" increases, the branches of the hyperbola are less "curved,"  and the focal points move further from the center.......(as expected)

Finally....here's a graph when b = 100........https://www.desmos.com/calculator/26r6fevyc3

Notice that the hyperbola appears to be almost "upright"  !!!

 

 Dec 14, 2014
 #1
avatar+118723 
+5
 Dec 14, 2014
 #2
avatar+130511 
+10
Best Answer

There are infinite possibilites for this....

The equation will be in the form

x^2 / 121  - y^2 / b^2   = 1        where "a" = 11 and a^2  = 121

Here's a graph when b = 5.........https://www.desmos.com/calculator/ptqorqve0j

Here's a graph when b = 13.........https://www.desmos.com/calculator/tmv8vqrcl6

As "a" stays constant and "b" increases, the branches of the hyperbola are less "curved,"  and the focal points move further from the center.......(as expected)

Finally....here's a graph when b = 100........https://www.desmos.com/calculator/26r6fevyc3

Notice that the hyperbola appears to be almost "upright"  !!!

 

CPhill Dec 14, 2014
 #3
avatar+118723 
+5

Thanks Chris :)

 Dec 14, 2014
 #4
avatar+130511 
0

OK.....

 

 Dec 14, 2014

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