document.write( "Question 615729: What is the equation of the asymptotes to the problem 16(x-2)^2-9(y+1)^2=-144 \n" ); document.write( "
Algebra.Com's Answer #388640 by lwsshak3(11628)\"\" \"About 
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What is the equation of the asymptotes to the problem
\n" ); document.write( "16(x-2)^2-9(y+1)^2=-144
\n" ); document.write( "divide by -144
\n" ); document.write( "-(x-2)^2/9+(y+1)^2/16=1
\n" ); document.write( "(y+1)^2/16-(x-2)^2/9=1
\n" ); document.write( "This is an equation of a hyperbola with vertical transverse axis.
\n" ); document.write( "Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center
\n" ); document.write( "For given equation:
\n" ); document.write( "center:(2,-1)
\n" ); document.write( "a^2=16
\n" ); document.write( "a=√16=4
\n" ); document.write( "b^2=9
\n" ); document.write( "b=√9=3
\n" ); document.write( "..
\n" ); document.write( "Asymptotes are straight lines that intersect at the center. Equation: y=mx+b, m=slope, b=y-intercept
\n" ); document.write( "Slopes of asymptotes with vertical transverse axis=±a/b=±4/3
\n" ); document.write( "Equation of asymptote with negative slope, -4/3
\n" ); document.write( "y=-4x/3+b
\n" ); document.write( "solve for b using coordinates of center (2,-1)
\n" ); document.write( "-1=-4*2/3+b
\n" ); document.write( "b=-1+8/3=5/3
\n" ); document.write( "equation: y=-4x/3+5/3
\n" ); document.write( "..
\n" ); document.write( "Equation of asymptote with positive slope, 4/3
\n" ); document.write( "y=4x/3+b
\n" ); document.write( "solve for b using coordinates of center (2,-1)
\n" ); document.write( "-1=4*2/3+b
\n" ); document.write( "b=-1-8/3=-11/3
\n" ); document.write( "equation: y=4x/3-11/3
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