document.write( "Question 616686: Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci.
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document.write( "(x-3)^2/9 - (y+5)^2/16 = 1 \n" );
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Algebra.Com's Answer #387846 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci. \n" ); document.write( "(x-3)^2/9 - (y+5)^2/16 = 1 \n" ); document.write( "** \n" ); document.write( "Equation is that of a hyperbola with vertical transverse axis. \n" ); document.write( "Its standard form: (y-k)^2/a^2-(x-h)^2/b^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( "vertices: (0, 0±a)=(0,0±4)=(0,-4) and (0,4) \n" ); document.write( ".. \n" ); document.write( "b^2=9 \n" ); document.write( "b=√9=3 \n" ); document.write( ".. \n" ); document.write( "c^2=a^2+b^2=16+9=25 \n" ); document.write( "c=√25=5 \n" ); document.write( "foci:(0, 0±c)=(0,0±5)=(0,-5) and (0,5) \n" ); document.write( "... \n" ); document.write( "slopes of asymptotes with vertical transverse axis=±a/b=±4/3 \n" ); document.write( "asymptotes are straight lines which intersect at the center. \n" ); document.write( "equation:y=mx+b, m=slope, b=y-intercept \n" ); document.write( "y-intercept=0, since asymptotes go thru the origin(0,0) \n" ); document.write( "equation of asymptote with negative slope: y=-4x/3 \n" ); document.write( "equation of asymptote with positive slope: y=4x/3 \n" ); document.write( ".. \n" ); document.write( "see graph below: \n" ); document.write( "y=±(16+16x^2/9)^.5\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |