SOLUTION: What is the equation of the asymptotes to the problem 16(x-2)^2-9(y+1)^2=-144
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Question 615729: What is the equation of the asymptotes to the problem 16(x-2)^2-9(y+1)^2=-144
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
What is the equation of the asymptotes to the problem
16(x-2)^2-9(y+1)^2=-144
divide by -144
-(x-2)^2/9+(y+1)^2/16=1
(y+1)^2/16-(x-2)^2/9=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation:
center:(2,-1)
a^2=16
a=√16=4
b^2=9
b=√9=3
..
Asymptotes are straight lines that intersect at the center. Equation: y=mx+b, m=slope, b=y-intercept
Slopes of asymptotes with vertical transverse axis=±a/b=±4/3
Equation of asymptote with negative slope, -4/3
y=-4x/3+b
solve for b using coordinates of center (2,-1)
-1=-4*2/3+b
b=-1+8/3=5/3
equation: y=-4x/3+5/3
..
Equation of asymptote with positive slope, 4/3
y=4x/3+b
solve for b using coordinates of center (2,-1)
-1=4*2/3+b
b=-1-8/3=-11/3
equation: y=4x/3-11/3
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