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With the hyperbolic equation 4x2 - y2 + 8x - 4y - 4 = 0, are the asymptotes y = (+,-) (4x + 4) - 2

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 Mar 11, 2016
 #1
avatar+128055 
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4x^2 - y^2 + 8x - 4y - 4 = 0

 

Complete the square on x and y

 

4[x^2 + 2x + 1]  -  [y^2 + 4y + 4]  =  4  + 4 -  4     factor and  simplify

 

4 ( x + 1)^2   - ( y + 2)^2    =  4       divide through by 4

 

(x + 1)^2  - (y + 2)^2 / 4  = 1      and we can put this in standard form as

 

(x + 1)^2  - (y + 2)^2   = 1

     1               4

 

The equation of the asymptotes is given by :

 

y = ± (b/a)(x - h) + k      and   b = 2,  a = 1,  h = -1    and k  = -2   so we have

 

y = ± (2/1) (x - -1) + (- 2)    which simplifies to

 

y = ± 2(x + 1) - 2

 

So the two two equations of the asymptotes further simplify to :

 

y = 2x   and y = -2x - 4

 

Here's the graph, Shades  : https://www.desmos.com/calculator/hguyqxzlql

 

 

cool cool cool

 Mar 11, 2016
 #2
avatar+466 
0

Ah, ok, I missed something before, thanks a lot! cool

 Mar 11, 2016

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