Question

xpatrat+4ypatrat-6x+4y+10< sau = 0

Answer 1

$x^{2} + 4y{2} -6x+4y+ 10 \leq0 \\\ \textless \ br /\ \textgreater \ (x^{2} -3*2*x + 9) +( (2y)^{2} + 2*1*2*y + 1) \leq 0 \\\ \textless \ br /\ \textgreater \ (x-3)^{2} + (2y+1)^{2} \leq 0 \\ \\ \ \textless \ br\\ \textgreater \ (x-3)^{2} = 0 =\ \textgreater \ x= 3 \\ \ \textless \ br /\ \textgreater \ (2y+1)^{2} = 0 =\ \textgreater \ y = \frac{-1}{2} \\ \\ \ \textless \ br /\ \textgreater \ S = ( 3; \frac{-1}{2})$