document.write( "Question 28101: What is the equation for the ellipse, the foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5)and (0,-5)? \n" ); document.write( "
Algebra.Com's Answer #15328 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! What is the equation for the ellipse, the foci are at (12,0) and (-12,0). The endpoints of the minor axis are at (0,5)and (0,-5)? \n" ); document.write( "-------------------------------------------------------- \n" ); document.write( "STANDARD EQN.OF ELLIPSE IS \n" ); document.write( "(X-H)^2/A^2 + (Y-K)^2/B^2 = 1 ,FOR A>B, WHOSE FOCI ARE (H+A*E,K)AND (H-A*E,K)WHERE \n" ); document.write( "E=SQRT{(A^2-B^2)/A^2} AND MINOR AXIS IS ALONG X=H \n" ); document.write( "AND LENGTH OF MAJOR AXIS =2A AND LENGTH OF MINOR AXIS =2B \r \n" ); document.write( "\n" ); document.write( "--------------------------------------- \n" ); document.write( "WE FIND X COORDINATES OF END POINTS OF MINOR AXIS ARE 0.HENCE MINOR AXIS IS Y AXIS. \n" ); document.write( "FURTHER WE FIND Y COORDINATES OF FOCI ARE 0...HENCE MAJOR AXIS IS X AXIS \n" ); document.write( "HENCE A>B \n" ); document.write( "HENCE CENTRE IS (0,0).SO H=K=0 \n" ); document.write( "LENGTH OF MINOR AXIS = 5+5 = 10 = 2B...OR B=5 \n" ); document.write( "SO NOW WE NEED TO FIND ONLY A TO GET THE EQN.OF ELLIPSE. \n" ); document.write( "WE GOT FOCI AS (12,0)=(0+A*E,0) \n" ); document.write( "AE=12....OR E=12/A \n" ); document.write( "BUT \n" ); document.write( "E=SQRT{(A^2-B^2)/A^2} =12/A...SQUARING... \n" ); document.write( "144/A^2=(A^2-B^2)/A^2 \n" ); document.write( "A^2-B^2=144 \n" ); document.write( "A^2=144+5^2=144+25=169 \n" ); document.write( "A=13 \n" ); document.write( "HENCE EQN. OF ELLIPSE IS \n" ); document.write( "X^2/169 + Y^2/25 = 1\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |