SOLUTION: Find the center, vertices, foci, and asymptotes of the hyperbola. {{{(x+4)^2/16 - y^2/16=1}}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the center, vertices, foci, and asymptotes of the hyperbola. {{{(x+4)^2/16 - y^2/16=1}}}      Log On


   

Question 813164: Find the center, vertices, foci, and asymptotes of the hyperbola.
%28x%2B4%29%5E2%2F16+-+y%5E2%2F16=1
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the center, vertices, foci, and asymptotes of the hyperbola.
%28x%2B4%29%5E2%2F16+-+y%5E2%2F16=1
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Given hyperbola has a horizontal transverse axis.
Its standard form of equation: %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of center.
For given hyperbola:
center: (-4,0)
a^2=16
a=√16=4
vertices:(-4±a,0)=(-4±4,0)=(-8,0) and (0,0)
..
b^2=16
b=√16=4
..
c^2=a^2+b^2=16+16=32
c=√32≈5.7
foci:(-4±c,0)=(-4±5.7,0)=(-9.7,0) and (1.7,0)
..
slope of asymptotes=±b/a=±4/4=±1
Equations of asymptotes:y=mx+b, m=slope,b=y-intercept
For slope=1
y=x+b
using coordinates of center to solve for b
0=-4+b
b=4
equation: y=x+4
..
For slope=-1
y=-x+b
using coordinates of center to solve for b
0=4+b
b=-4
equation: y=-x-4