document.write( "Question 454063: What are the equations of the asymptotes and the coordinates of the foci for this equation \"+36x%5E2-y%5E2-4y=-32+\" \n" ); document.write( "
Algebra.Com's Answer #311920 by lwsshak3(11628)\"\" \"About 
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What are the equations of the asymptotes and the coordinates of the foci for this equation 36x^2-y^2-4y=-32
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\n" ); document.write( "36x^2-y^2-4y=-32
\n" ); document.write( "36x^2-(y^2-4y+4)=-32-4=-36
\n" ); document.write( "36x^2-(y-2)^2=-36
\n" ); document.write( "divide by -36
\n" ); document.write( "-x^2+(y-2)^2/36=1
\n" ); document.write( "(y-2)^2/36-x^2=1
\n" ); document.write( "This is a hyperbola with vertical transverse axis and center at (0,2)
\n" ); document.write( "a^2=36
\n" ); document.write( "a=6
\n" ); document.write( "b^2=1
\n" ); document.write( "b=1
\n" ); document.write( "c^2=a^2+b^2=36+1=37
\n" ); document.write( "c=√37
\n" ); document.write( "coordinates of foci (0,2±√37
\n" ); document.write( "Equations of asymptotes: y=mx+b)=±(a/b)x+2=±6x+2
\n" ); document.write( "See graph as a visual check on the answers above
\n" ); document.write( "..
\n" ); document.write( "y=(36+36x^2)^.5+2\r
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