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A hyperbola has one of its foci at (3, 2), and the vertex of the hyperbola closer to this focus is at (4, 2). One of the asymptotes of the hyperbola has slope sqrt(2). Find the x-coordinate of the center of the hyperbola.

 Jul 15, 2022
 #1
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This hyperbola opens sideways...so.....let the center  be    at ( x , 2)

 

"c"  =  x -  3       →   c^2  =  x^2 - 6x + 9

"a"  =  x  - 4       →   a^2 =   x^2  - 8x + 16

 

"b"  = sqrt  [ c^2 - a^2 ]  =   sqrt [  (x^2 - 6x + 9) - (x^2 - 8x + 16)  ]  =  sqrt [ 2x - 7]  

 

The slope is given as

 

b / a   =  sqrt 2

 

sqrt [ 2x  -7 ]  / [x - 4 ]   =  sqrt 2

 

sqrt [ 2x - 7 ]  =  sqrt 2 *  [ x -4]           square both sides

 

2x - 7  = 2  [ x^2 - 8x + 16 ]

 

2x^2 - 18x + 39  =  0

 

          18  +√ [ 18^2 - 4*2 * 39 ]               18 +√ [ 12 ]               9 + √ [ 3 ]

x =    ______________________  =    ___________  =       ___________  ≈   5.366

                      2 * 2                                         4                               2

 

 

cool cool cool

 Jul 15, 2022

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