document.write( "Question 391623: How do I identify the equation (name the conic section), and then convert the equation to its standard form by completing the square on x and y
\n" ); document.write( "for the equation
\n" ); document.write( "4y^2-9x^2-16y-36x-56=0 \n" ); document.write( "
Algebra.Com's Answer #277990 by lwsshak3(11628)\"\" \"About 
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given:4y^2-9x^2-16y-36x-56=0
\n" ); document.write( "4y^2-16y-9x^2-36x -56 =0
\n" ); document.write( "4(y^2-4y+4)-9(x^2-4x+4) =56+16-36
\n" ); document.write( "4(y-2)^2-9(x+2)^2=36
\n" ); document.write( "÷36
\n" ); document.write( "(y-2)^2/9 - (x+2)^2/4 =1
\n" ); document.write( "This is a hyperbola with its center at (-2,2), and transverse axis vertical.
\n" ); document.write( "standard form of hyperbola, (y-k)^2/a^2-(x-h)^2 =1\r
\n" ); document.write( "\n" ); document.write( "a^2 =9
\n" ); document.write( "a=3
\n" ); document.write( "b^2=4
\n" ); document.write( "b=2\r
\n" ); document.write( "\n" ); document.write( "c=sqrt(a^2)+(b^2
\n" ); document.write( "sqrt(13)
\n" ); document.write( "length of transverse axis = 2a =2*3=6
\n" ); document.write( "length of conjugate axis=2b=2*2=4
\n" ); document.write( "vertices=(-2,2±a)=(2,2±3)
\n" ); document.write( "foci =(-2,2±c)=(2±,sqrt(13)
\n" ); document.write( "asymptotes = (3/2)x+5 and (-3/2)x-1
\n" ); document.write( "see graph below\r
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