area of triangle

x²-y²+2y=1

x²=(y-1)²

so x=y-1
and x+y-1=0

Their angular bisectors are not tough to find... (by a graph u can do it even quicker..)

one will be x=0 another will be y=1

now plot the graph .. it will be a right angled triangle..

points (0,1) (0,3) and (2,1)

Area will be "removed"3"end edit" 2

yeah saket.. u din see the last step..

there was a small mistake.. the area was indeed 2..

I just missed that one in jaldi baji :) ;)

first we find the slopes of the two lines by taking

ax²+mxy+by²=0

The two roots of y/x will give m1 and m2

so if we get the slopes as m1 and m2

we take the equations as

(y-m1x-c1)(y-m2x-c2)=0

and then we compare the coefficients....

A dirty method.. but in the most general case, this is what we need to do

I think there is one.. i dont remember it right now.. let me think if i can figure it out or else search it on the net...