document.write( "Question 874125: determine the center, vertices, foci and equations of the asymptotes \r
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document.write( "(X-5)^2 OVER 4 - (Y+2)^2 OVER 16 =1 \r
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Algebra.Com's Answer #527381 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! determine the center, vertices, foci and equations of the asymptotes \n" ); document.write( "(X-5)^2 OVER 4 - (Y+2)^2 OVER 16 =1 \n" ); document.write( "*** \n" ); document.write( "(x-5)^2/4-(y+2)^2/16=1 \n" ); document.write( "This is an equation of a hyperbola with vertical transverse axis. \n" ); document.write( "Its standard form of equation: ((x-h)^2/b^2-(y-k)^2/a^2=1}}}, (h,k)=coordinates of center \n" ); document.write( "For given hyperbola: \n" ); document.write( "center: (5-2) \n" ); document.write( "a^2=16 \n" ); document.write( "a=4 \n" ); document.write( "vertices: (5±a,-2)=(5±4,-2)=(-1,-2) and (9,-2) \n" ); document.write( "b^2=4 \n" ); document.write( "b=2 \n" ); document.write( "c^2=a^2+b^216+4=20 \n" ); document.write( "c=√20≈4.5 \n" ); document.write( "foci: (5±c,-2)=(5±4.5,-2)=(-1.5,-2) and (9.5,-2) \n" ); document.write( ".. \n" ); document.write( "The two asymptotes are straight line equations that go through the center (5,-2), and take the form y=mx+b, m=slope, b=y-intercept. \n" ); document.write( "slopes of asymptotes for hyperbolas with vertical transverse axis=±a/b=±4/2=±2 \n" ); document.write( ".. \n" ); document.write( "For asymptote with negative slope: \n" ); document.write( "y=-2x+b \n" ); document.write( "solve for b using coordinates of the center \n" ); document.write( "-2=-2*5+b \n" ); document.write( "b=8 \n" ); document.write( "equation:y=-2x+8 \n" ); document.write( ".. \n" ); document.write( "For asymptote with positive slope: \n" ); document.write( "y=2x+b \n" ); document.write( "solve for b using coordinates of the center \n" ); document.write( "-2=2*5+b \n" ); document.write( "b=-12 \n" ); document.write( "equation:y=2x-12\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |