document.write( "Question 743731: Find the equation of the hyperbola that the transverse axis is parallel to the x-axis, center at (2,-2), passing through (2+3√2,0) and (2+3√10,4). \n" ); document.write( "
Algebra.Com's Answer #453037 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! Find the equation of the hyperbola that the transverse axis is parallel to the x-axis, center at (2,-2), passing through (2+3√2,0) and (2+3√10,4). \n" ); document.write( "*** \n" ); document.write( "This is a hyperbola with horizontal transverse axis. \n" ); document.write( "Its standard form of equation: \n" ); document.write( "For given hyperbola: \n" ); document.write( "Given center: (2,-2) \n" ); document.write( "Equation: \n" ); document.write( "using given coordinates to solve for a and b: \n" ); document.write( "(2+3√2-2)^2/a^2-(0+2)^2/b^2=1 \n" ); document.write( "(2+3√10-2)^2/a^2-(4+2)^2/b^2=1 \n" ); document.write( ".. \n" ); document.write( "18/a^2-4/b^2=1 \n" ); document.write( "90/a^2-36/b^2=1 \n" ); document.write( "let x=1/a^2 \n" ); document.write( "let y=1/b^2 \n" ); document.write( ".. \n" ); document.write( "18x-4y=1 \n" ); document.write( "90x-36y=1 \n" ); document.write( ".. \n" ); document.write( "-162x+36y=-9 \n" ); document.write( "90x-36y=1 \n" ); document.write( "add \n" ); document.write( "-72x=-8 \n" ); document.write( "x=8/72=1/9=1/a^2 \n" ); document.write( "a^2=9 \n" ); document.write( ".. \n" ); document.write( "4y=18x-1=2-1=1 \n" ); document.write( "y=1/4=1/b^2 \n" ); document.write( "b^2=4 \n" ); document.write( "Equation of given hyperbola: \n" ); document.write( " \n" ); document.write( ".. \n" ); document.write( " |