Question 479582: y^(2)/(16)-x^(2)/(9)=1
Identify the curve, find the center,asymptotes, foci; then sketch the curve
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! y^(2)/(16)-x^(2)/(9)=1
Identify the curve, find the center,asymptotes, foci; then sketch the curve
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y^2/16-x^2/9=1
This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1
for given equation:
Center(0,0)
a^2=16
a=4
b^2=9
b=3
c^2=a^2+b^2=16+9=25
c=√25=5
Foci: (0,0±c)=(0±0±5)=(0,5) and (0,-5)
Asymptotes:
slope, m=±a/b=±4/3
Equation: y=±mx+b
asymptotes go thru center, so b=0
equations:y=±4x/3
see graph below as a visual check on answers
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y=±(16+16x^2/9)^.5
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