Question

The values of $ x $ and $ y $ such that $ y $ satisfy the equation $ (x,\,\,\,y\,\,\,\in $ real numbers) $ {{x}^{2}}-xy+{{y}^{2}}-4x-4y+16=0 $ is

• A. $ 4,\,\,4 $
• B. $ 3,\,\,3 $
• C. $ 2,\,\,2 $
• D. none of these

Answer 1

Answer: (A)

Given, $ {{x}^{2}}-xy+{{y}^{2}}-4x-4y+16=0 $
$ \Rightarrow $ $ {{x}^{2}}-(y+4)x+{{y}^{2}}-4y+16=0 $
For real $ x,\,\,{{(y+4)}^{2}}-4({{y}^{2}}-4y+16)\ge 0 $
$ \Rightarrow $ $ -3{{y}^{2}}+24y-48=0 $
$ \Rightarrow $ $ {{y}^{2}}-8y+16=0 $
$ \Rightarrow $ $ {{(y-4)}^{2}}=0\Rightarrow y=4 $
$ \therefore $ $ x=4 $ $ \Rightarrow $ $ (x,\,\,y)=(4,\,\,4) $