1
2
0
-1
We have,
\(x ^2 +(x−2y−1) ^2 =−4y(x+y) \)
\(⇒ x^ 2 +4xy+4y^ 2 +(x−2y−1)^ 2 =0\)
\(⇒ (x+2y) ^2 +(x−2y−1) ^2 =0\)
Since squares cannot be negative, all of the square terms in the equation must be zero for the L.H.S. to be 0.
\(x - 2y - 1 = 0 \)
\(⇒ x - 2y = 1\)