SOLUTION: use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3

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Question 715292: use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3
Answer by lwsshak3(11628) About Me  (Show Source):
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use the eccentricity of each hyperbola to find its equation in standard form center(4,1),horizontal transverse axis is 12 and eccentricity 4/3
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Standard form of equation for a hyperbola with horizontal transverse axis:
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of center.
For given hyperbola:
center: (4,1)
length of transverse axis=12=2a
a=6
a^2=36
eccentricity:=4/3=c/a
c=4a/3=24/3=8
c^2=64
c^2=a^2+b^2
b^2=c^2-a^2=64-36=28
Equation of given hyperbola:
%28x-4%29%5E2%2F36-%28y-1%29%5E2%2F28=1