document.write( "Question 39217: Write the equation of a hyperbola from the given information. Graph the equation. Place the center of the hyperbola at the orgin of the coordinate plane. One focus is located at (0,6); one vertex at (0,-(square root) 7 \n" ); document.write( "
Algebra.Com's Answer #24660 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! Write the equation of a hyperbola from the given information. Graph the equation. Place the center of the hyperbola at the orgin of the coordinate plane. One focus is located at (0,6); one vertex at (0,-(square root) 7 \n" ); document.write( "------------------------------------------------------------------------\r \n" ); document.write( "\n" ); document.write( "THE EQN. OF A HYPERBOLA IN STD.FORM IS \n" ); document.write( "(X-H)^2/A^2 - (Y-K)^2/B^2=-1…. \n" ); document.write( "WHERE \n" ); document.write( "(H,K) IS CENTRE…HERE WE HAVE CENTRE IS (0,0).SO H=K=0 \n" ); document.write( "VERTICES ARE (0,B),(0,-B)...B=SQRT(7) \n" ); document.write( "FOCI ARE (H,K+BE)AND (H,K-BE),THAT IS (0,BE),(0,-BE)....THEY ARE (0,6) AND (0,-6) SINCE CENTRE IS ORIGIN.HENCE BE=6.......E=6/SQRT(7) \n" ); document.write( "E^2=36/7=(A^2+B^2)/B^2=(A^2+7)/7 \n" ); document.write( "A^2+7=36....A^2=29 \n" ); document.write( "HENCE EQN.IS \n" ); document.write( "X^2/29-Y^2/7=-1...OR....Y^2/7-X^2/29=1 \n" ); document.write( " |