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Find the asymptote of -4x^2 +y^2 -2y = 3 with a postive slope. Answer should be in y=mx+b form .

 

I was gettiing confused when trying to simplify the equation in to the equation of a hyperbola.

 Dec 8, 2018
 #1
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-4x^2 +y^2 -2y = 3      rewrite as

 

y^2 - 2y - 4x^2 = 3        complete the square on y

 

y^2 - 2y + 1 - 4x^2 = 3 + 1

 

(y - 1)^2  -   4x^2 =   4         divide both sides by 4

 

(y - 1)^2     -   x^2  

______         ____  =      1

   4                  1

 

In this form,   a = 2  , b = 1     h = 0    k = 1

 

The equation of the asymptote with the positive slope is 

 

y =  (a/b) ( x - h) + k        ....so...

 

y = (2/1) ( x - 0) + 1

 

y = 2x + 1

 

Here's the graph :   https://www.desmos.com/calculator/tenipf6zfq

 

 

 

cool cool cool

 Dec 8, 2018

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