document.write( "Question 618962:  Find the equations of the asymptotes of the hyperbola.\r
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document.write( "25y^2 - 16x -400 = 0 \n" );
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| Algebra.Com's Answer #389204 by lwsshak3(11628)      You can put this solution on YOUR website! Find the equations of the asymptotes of the hyperbola. \n" ); document.write( "25y^2 - 16x^2 -400 = 0 \n" ); document.write( "25y^2-16x^2=400 \n" ); document.write( "Divide by 400 \n" ); document.write( "y^2/16-x^2/25=1 \n" ); document.write( "This is an equation of a hyperbola with vertical transverse axis \n" ); document.write( "Its form of equation: (y-k)^2/a^2-(x-h)^2=1, (h,k)=(x,y) coordinates of the center. \n" ); document.write( "For given equation: \n" ); document.write( "center: (0,0) \n" ); document.write( "a^2=16 \n" ); document.write( "a=√16=4 \n" ); document.write( "b^2=25 \n" ); document.write( "b=√25=5 \n" ); document.write( ".. \n" ); document.write( "Asymptotes are straight lines that intersect at the center(0,0). Equation: y=mx+b, m=slope, b=y intercept \n" ); document.write( "Slopes of asymptotes for hyperbolas with vertical transverse axis=a/b=4/5 \n" ); document.write( "Equation of asymptote with slope<0 \n" ); document.write( "y=-4x/5+b \n" ); document.write( "since asymptote go thru center, y-intercept, b=0 \n" ); document.write( "so, equation: y=-4x/5 \n" ); document.write( ".. \n" ); document.write( "By the same reasoning, \n" ); document.write( "Equation of asymptote with slope>0 \n" ); document.write( "y=4x/5 \n" ); document.write( " \n" ); document.write( " |