document.write( "Question 732626: If the equation for a hyperbola is not in the standard form already, how do you change it so you can graph it? I have this equation: 6(x-3)^2-4(y+1)^2=96. Is it necessary to put it in the standard hyperbola form? If so, how would you do this? \n" ); document.write( "
Algebra.Com's Answer #448061 by lwsshak3(11628)\"\" \"About 
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If the equation for a hyperbola is not in the standard form already, how do you change it so you can graph it? I have this equation: 6(x-3)^2-4(y+1)^2=96. Is it necessary to put it in the standard hyperbola form? If so, how would you do this?
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\n" ); document.write( "Standard forms of hyperbola:
\n" ); document.write( "For hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
\n" ); document.write( "For hyperbola with vertical transverse axis: \"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\", (h,k)=(x,y) coordinates of center
\n" ); document.write( "..
\n" ); document.write( "For given equation of hyperbola:
\n" ); document.write( "\"6%28x-3%29%5E2-4%28y%2B1%29%5E2=96\"
\n" ); document.write( "change to standard form:
\n" ); document.write( "divide both sides by 96
\n" ); document.write( "\"%28x-3%29%5E2%2F16-%28y%2B1%29%5E2%2F24=1\"
\n" ); document.write( "center: (3,-1)
\n" ); document.write( "a^2=16
\n" ); document.write( "a=4
\n" ); document.write( "length of horizontal transverse axis=2a=8
\n" ); document.write( "..
\n" ); document.write( "b^2=24
\n" ); document.write( "b=√24
\n" ); document.write( "length of conjugate axis=2b=2√24=2√6
\n" ); document.write( "..
\n" ); document.write( "slopes of the asymptotes=±b/a=2±√6/4=±√6/2
\n" ); document.write( "Equation of asymptote with negative slope:
\n" ); document.write( "y=mx+b=-√6x/2+b
\n" ); document.write( "solve for b using coordinates of center which are on the asymptote line.
\n" ); document.write( "-1=-√6*3/2+b
\n" ); document.write( "b=-1+3.6742=2.6742
\n" ); document.write( "Equation:y=-√6x/2+2.6742
\n" ); document.write( "..
\n" ); document.write( "Equation of asymptote with positive slope:
\n" ); document.write( "y=mx+b=√6x/2+b
\n" ); document.write( "solve for b using coordinates of center which are on the asymptote line.
\n" ); document.write( "-1=√6*3/2+b
\n" ); document.write( "b=-1-3.6742=-4.6742
\n" ); document.write( "Equation:y=√6x/2-4.6742
\n" ); document.write( "..
\n" ); document.write( "To graph the hyperbola manually, it is best to draw a rectangle around the center:
\n" ); document.write( "draw a horizontal line thru the center (3,-1) with end points 4 units from center. (horizontal transverse axis)
\n" ); document.write( "draw a vertical line thru the center (3,-1) with end points√24/2 units from center.(conjugate axis)
\n" ); document.write( "draw lines thru the end points to form a rectangle
\n" ); document.write( "Asymptotes go thru the corners of this rectangle and the center.
\n" ); document.write( "You now should be able to graph the hyperbola knowing the coordinates of the center, asymptotes, and the fact that hyperbola has a horizontal transverse axis
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