SOLUTION: how do you find the vertices, co-vertices, foci, and asymptotes of: 9(x+2)^2 -16(y-1)^2=-144

Question 615541: how do you find the vertices, co-vertices, foci, and asymptotes of:
9(x+2)^2 -16(y-1)^2=-144
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
how do you find the vertices, co-vertices, foci, and asymptotes of:
9(x+2)^2 -16(y-1)^2=-144
divide by -144
-(x+2)^2/16+(y-1)^2/9=1
(y-1)^2/9-(x+2)^2/16=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: (y-1)^2/9-(x+2)^2/16=1
center: (-2,1)
a^2=9
a=3
vertices:(-2,1�a)=(-2,1�3)=(-2,-2) and (-2,4)
..
b^2=16
b=4
co-vertices:(-2�b,1)=(-2�4,1)=(-6,1) and (2,1)
..
c^2=a^2+b^2=9+16=25
c=√25=5
foci::(-2,1�c)=(-2,1�5)=(-2,-4) and (-2,6)
..
Asymptotes are straight lines that intersect at the center
slopes of asymptotes=�a/b=�3/4
..
Equation of asymptote with negative slope
y=mx+b, m=slope, b=intercept
y=-3x/4+b
Solve for b using coordinates of center
1=-3*-2/4+b
b=1-6/4=-2/4=-1/2
Equation: y=-3x/4-1/2
..
Equation of asymptote with positive slope
y=mx+b, m=slope, b=intercept
y=3x/4+b
Solve for b using coordinates of center
1=3*-2/4+b
b=1+6/4=10/4=5/2
Equation: y=3x/4+5/2

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