The number of pairs of real numbers $(x,y)$ that satisfy $2x^2+y^2+2xy-2y+2=0$

Question

$2x^2+y^2+2xy-2y+2=0$

=$(x+y)^2+x^2-2y+1$

I thought that this could be expressed in the form of ellipse but I failed due to $2xy$ term

Then I used calculus

$f(x,y)=2x^2+y^2+2xy-2y+2=0 $

$\implies$ $f_x= 4x+2y=0................(1)$

[Here $f_x \implies \frac{d(fx,y)}{dx}$]

& $f_y=2y+2x-2=0................(2)$

[Here $f_y \implies \frac{d(fx,y)}{dy}$]

From $(1)$&$(2)$, we get $x=-1,y=2$

Also $f_{xx}(-1,2)=4\space,\space f_{yy}(-1,2)=2\space,\space f_{xy}(-1,2)=2$

[$f_{xx}\implies \frac{d^2(f(x,y))}{dx^2} \space f_{xy}\implies \frac{d^2(f(x,y))}{dx\cdot dy} $]

$\therefore\space(f_{xx}\cdot f_{yy}\space -\space f_{xy}^2)=8-2=6>0$

$\therefore$ We get a minima at $(-1,2)$

$\therefore$ # of pairs satisfying the reqd. equation is 1

Is my approach and conclusion right? If not please give me the correct approach and conclusion citing the proper reason.

Answer 1

HInt:

$$2x^2+x(2y)+y^2-2y+2=0$$

As $x$ is real, the discriminant must be $\ge0$

$$(2y)^2-8(y^2-2y+2)\ge0\iff0\ge-4(y-2)^2\iff(y-2)^2\le0$$

But as $y$ is real, $(y-2)^2\ge0$

Answer 2

$$2x^2+y^2+2xy-2y+2=\frac{(2x+y)^2+(y-2)^2}{2}=0$$

if and only if

$$2x+y=y-2=0$$