document.write( "Question 841004: WHAT IS THE FOCI FOR 4X^2-9Y^2=8x+54-41=0
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Algebra.Com's Answer #506750 by lwsshak3(11628) $\"\"$ $\"About$

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WHAT IS THE FOCI FOR
\n" ); document.write( "4X^2-9Y^2=8x+54-41=0
\n" ); document.write( "4x^2-8x-9y^2=-13
\n" ); document.write( "complete the square
\n" ); document.write( "4(x^2-2x+1)-9y^2=-13+4
\n" ); document.write( "4(x-1)^2-9y^2=-9
\n" ); document.write( "divide by -9
\n" ); document.write( " $\"y%5E2-%28x-1%29%5E2%2F%289%2F4%29=1\"$
\n" ); document.write( "This is an equation of a hyperbola with vertical transverse axis.
\n" ); document.write( "Its standard form: $\"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\"$ , (h,k)=(x,y) coordinates of center.
\n" ); document.write( "for given hyperbola:
\n" ); document.write( "center:(1,0)
\n" ); document.write( "a^2=1
\n" ); document.write( "a=1
\n" ); document.write( "b^2=9/4
\n" ); document.write( "b=3/2
\n" ); document.write( "c^2=a^2+b^2=1+9/4=13/4=3.25
\n" ); document.write( "Foci=(1,0ąc),(1,ą3.25)=(1,-3.25) and (1+3.25)
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