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

Question 841004: WHAT IS THE FOCI FOR 4X^2-9Y^2=8x+54-41=0

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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

This is an equation of a hyperbola with vertical transverse axis.
Its standard form: , (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)

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