document.write( "Question 631670: where is the asymptotes, center, and equations of asymptotes. for this equation:\r
\n" ); document.write( "\n" ); document.write( "4y^2 - 16y - x^2 + 2x = 10 \n" ); document.write( "
Algebra.Com's Answer #397748 by lwsshak3(11628)\"\" \"About 
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where is the asymptotes, center, and equations of asymptotes. for this equation:
\n" ); document.write( "4y^2 - 16y - x^2 + 2x = 10
\n" ); document.write( "complete the square
\n" ); document.write( "4(y^2-4y+4)-(x^2-2x+1) = 10+16-1
\n" ); document.write( "4(y-2)-(x-1)=25
\n" ); document.write( "\"%28y-2%29%2F%2825%2F4%29-%28x-1%29%2F25=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\"
\n" ); document.write( "For given equation:
\n" ); document.write( "center: (1,2)
\n" ); document.write( "a^2=25/4
\n" ); document.write( "a=√(25/4)=5/2
\n" ); document.write( "b^2=25
\n" ); document.write( "b=√25=5
\n" ); document.write( "asymptotes are straight lines that go thru the center: equation: y=mx+b, m=slope, b=y-intercept
\n" ); document.write( "slopes, m, of asymptotes for hyperbolas with vertical transverse axis=±a/b=±(5/2)5=±5/10=±1/2
\n" ); document.write( "..
\n" ); document.write( "asymptote with negative slope: y=-x/2+b
\n" ); document.write( "solve for b using coordinates of center (1,2)
\n" ); document.write( "2=-1/2+b
\n" ); document.write( "b=5/2
\n" ); document.write( "equation: y=-x/2+5/2
\n" ); document.write( "..
\n" ); document.write( "asymptote with positive slope: y=x/2+b
\n" ); document.write( "solve for b using coordinates of center (1,2)
\n" ); document.write( "2=1/2+b
\n" ); document.write( "b= 3/2
\n" ); document.write( "equation: y=-x/2+3/2
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