Question 596617
write the equation for the following asymptotes for the following hyperbola x^2-4y^2=16
divide by 16
x^2/16-y^2/4=1
This is an equation for a hyperbola with horizontal transverse axis of the standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center.
For given hyperbola:
center:(0,0)
a^2=16
a=√16=4
..
b^2=4
b=2
..
slopes of asymptotes, m: ±b/a=±2/4=±1/2
equations of asymptotes:
y=mx+b and y=-mx+b, m=slope, b=y-intercept
y=x/2+b and y=-x/2+b
since asymptotes go thru origin, (0,0), y-intercept,b=0
equations therefore are:
y=x/2 and y=-x/2