Question 874125: determine the center, vertices, foci and equations of the asymptotes
(X-5)^2 OVER 4 - (Y+2)^2 OVER 16 =1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! determine the center, vertices, foci and equations of the asymptotes
(X-5)^2 OVER 4 - (Y+2)^2 OVER 16 =1
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(x-5)^2/4-(y+2)^2/16=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form of equation: ((x-h)^2/b^2-(y-k)^2/a^2=1}}}, (h,k)=coordinates of center
For given hyperbola:
center: (5-2)
a^2=16
a=4
vertices: (5±a,-2)=(5±4,-2)=(-1,-2) and (9,-2)
b^2=4
b=2
c^2=a^2+b^216+4=20
c=√20≈4.5
foci: (5±c,-2)=(5±4.5,-2)=(-1.5,-2) and (9.5,-2)
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The two asymptotes are straight line equations that go through the center (5,-2), and take the form y=mx+b, m=slope, b=y-intercept.
slopes of asymptotes for hyperbolas with vertical transverse axis=±a/b=±4/2=±2
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For asymptote with negative slope:
y=-2x+b
solve for b using coordinates of the center
-2=-2*5+b
b=8
equation:y=-2x+8
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For asymptote with positive slope:
y=2x+b
solve for b using coordinates of the center
-2=2*5+b
b=-12
equation:y=2x-12
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