Question 413360
i need to find the asymptotes of:
(x/4)2-(y/2)2=1

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Standard form of given hyperbola is: x^2/a^2-y^2/b^2=1
What you have here is: 
x^2/16-y^2/4=1
This is a hyperbola with center at (0,0) that open sideways, that is, its transverse axis is horizontal.
a^2=16
a=4
b^2=4
b=2
like you said, the equations of the asymptotes are: y=b/ax and-b/ax.
For the given hyperbola, y=(2/4)x and y=-(2/4)x
See the graph below for a visual picture of the asymptotes. You might note that equations of the asymptotes are equations of straight lines of standard form, y=mx+b, with the y-intercept=0 and the slopes, m=b/a and m=-b/a
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y=((x^2-16)/4)^.5
{{{ graph( 300, 300, -10, 10, -10, 10, ((x^2-16)/4)^.5,-((x^2-16)/4)^.5,.5x,-.5x) }}}