|
Question 818627: fine the asymptotes of the hyperbola given by x^2/16-y^2/36=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the asymptotes of the hyperbola given by
x^2/16-y^2/36=1
Given equation is that of a hyperbola with horizontal transverse axis and center at the origin (0,0)
Asymptotes are straight lines that go through the center and take the standard form: y=mx+b, m=slope, b=y-intercept
slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a
For given hyperbola:
a^2=16
a=√16=4
b^2=36
b=√36=6
slopes of asymptotes=±b/a=±6/4=±3/2
equation of asymptote with positive slope:
y=mx+b
y=3x/2+b
since asymptote goes thru center (0,0), b=0
so, y=3x/2
..
equation of asymptote with negative slope:
y=-3x/2
|
|
|
| |
