document.write( "Question 721973: Find the center, vertices, and foci of the hyperbola and put the equation in standard form (x^2)-(9y^2)+2x-54y-107=0 \n" ); document.write( "
Algebra.Com's Answer #442615 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Find the center, vertices, and foci of the hyperbola and put the equation in standard form \n" ); document.write( "(x^2)-(9y^2)+2x-54y-107=0 \n" ); document.write( "*** \n" ); document.write( "complete the square: \n" ); document.write( "(x^2)-(9y^2)+2x-54y-107=0 \n" ); document.write( "(x^2)+2x-(9y^2)-54y-107=0 \n" ); document.write( "(x^2+2x+1)-9(y^2+6y+9)=107+1+-81 \n" ); document.write( "(x+1)^2-9(y+3)^2=27 \n" ); document.write( " \n" ); document.write( "This is an equation of a hyperbola with horizontal transverse axis. \n" ); document.write( "Its standard form: \n" ); document.write( "For given hyperbola: \n" ); document.write( "center: (-1,-3) \n" ); document.write( "a^2=27 \n" ); document.write( "a=√27≈5.2 \n" ); document.write( "vertices: (-1±a,-3)=(-1±5.2,-3)=(-6.2,-3) and (4.2,-3) \n" ); document.write( ".. \n" ); document.write( "c^2=a^2+b^2=27+3=30 \n" ); document.write( "c=√30≈5.5 \n" ); document.write( "foci:(-1±c,-3)=(-1±5.5,-3)=(-6.5,-3) and (4.5,-3) \n" ); document.write( " |