X2 - Y2 + 4X = 0
IF WE SOLVE THE EQN ,WE GET THE EQN OF A HYPERBOLA WITH ECCENTRICITY = 0.
CAN ANY ONE EXPLAIN IF A HYPERBOLA CAN HAVE ECCEN.= 0.
IF IT IS POSSIBLE CAN YOU SHOW IT GRAPHICALLY
straight line
use condition for equation ax2+by2+2hxy+2gx+2fy+c=0
to represent a pair of lines
abc+2hgf-af2-bg2-ch2=0
this condition is satisfied hence it represents a pair of lines and not a HYPERBOLA of zero eccentricity.
thankx for showing interest
ACTUALLY ,I HAD BEEN WRONG IN CALCULATING THE ECCENTRICITY WHICH ACTUALLY IS SQUR ROOT OF 2 AND NOT ZERO
OTHERWISE IT IS AN EQUATION OF AN HYPERBOLA ONLY
i am pretty confused
can u send me the soln???
thanx in advance
I think it is not a continous line but two rays moving outward.
Mohit you may consider that a normal hyperbola arms are compressed to x - axis.
The dark line is your curve.
HEY MOHIT ! THE GIVEN EQUATION REPRESENTS A RECTANGULAR HYPERBOLA OF ECCENTRICITY SQ. ROOT OF 2 .