SOLUTION: WHAT IS THE FOCI FOR 4X^2-9Y^2=8x+54-41=0

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Question 841004: WHAT IS THE FOCI FOR 4X^2-9Y^2=8x+54-41=0
Answer by lwsshak3(11628) About Me  (Show Source):
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WHAT IS THE FOCI FOR
4X^2-9Y^2=8x+54-41=0
4x^2-8x-9y^2=-13
complete the square
4(x^2-2x+1)-9y^2=-13+4
4(x-1)^2-9y^2=-9
divide by -9
y%5E2-%28x-1%29%5E2%2F%289%2F4%29=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of center.
for given hyperbola:
center:(1,0)
a^2=1
a=1
b^2=9/4
b=3/2
c^2=a^2+b^2=1+9/4=13/4=3.25
Foci=(1,0±c),(1,±3.25)=(1,-3.25) and (1+3.25)