Daily Graph 08-08-09

i mean it wont be a curve or anything....just a set of points......am i right??

its the fast and the furious site..;-)

my god everything here is happening at lightning fast speed ... :D

got it..:)

@phillip: nice one ...

@honey: so the point will be (-1,-1) which is the only point satisfying the second curve as the sum of three squares is zero implying that individual terms are zero

wow philip.......

so for that expression to be zero...
x=y ,x=-1,y =-1

=> (-1,-1)

i think phillip has already written what nishant sir was asking for.........:-)

\frac{[(x-y)^2+(x+1)^2+(y+1)^2]}{2}

sorry sir...it was just a thought.....i will delete it

tooooooo much working
too complex graph.....will post it as soon as i draw it....

there will be only two points (x2 + y2 + x+ y +1-xy)=0
that at x=-1
and y=-1

no uttara..
here a=1,b=1,h=-1/2 =>ab-h²>0 => ellipse

what integer soln ?????

we have to sketch whole graph...............

should I unhide my ans now ????since it is confirmed it is correct ??

x³ + y³ + 3xy=1
(x+y-1)(x² + y² + x+ y +1-xy)=0
so eure is right about the x+y-1 line
:-)