Discussion

Explanation:

Given, x2 + (x – 2y - 1)2 = -4y(x + y)

⇒ x2 + (x – 2y - 1)2 = -4yx - 4y2 

⇒ x2 + 4yx + 4y2 + (x – 2y - 1)2 = 0

⇒ (x2 + 2 × x × 2y + (2y)2) + (x – 2y - 1)2 = 0

⇒ (x + 2y)2 + (x – 2y - 1)2 = 0

Sum of squares of two number can be 0 only when both the numbers are 0.

∴ x + 2y = 0 and x – 2y – 1 = 0

⇒ x – 2y = 1

Hence, option (a).

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