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..;-)

Yes so the remaining point will be (-1,-1)

@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

itz also valid if u put y=2
so thr wld be more pts

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

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

integer solution are x=-1 and y=-1

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
:-)