Q. The values of $x\ and\ y$ such that $y$ satisfy the equation $(x,y \in$ real number ) $x^{2} - xy + y^{2} - 4x - 4y + 16 = 0$ is
A) 4, 4
B) 3, 3
C) 2, 2
D) None of these
Explanation
(a)
Given, $x^{2} - xy + y^{2} - 4x - 4y + 16 = 0$
$\Rightarrow x^{2} - (y + 4)x + y^{2} - 4y + 16 = 0$
For real $x,(y + 4)^{2} - 4(y^{2} - 4y + 16) \geq 0$
$\Rightarrow \ - 3y^{2} + 24y - 48 = 0$
$\Rightarrow y^{2} - 8y + 16 = 0$
$\Rightarrow (y - 4)^{2} = 0$
$\Rightarrow y = 4$
$\therefore x = 4$
$\Rightarrow (x,y) = (4,\ 4)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.