Answer
ellipse
Work Step by Step
The general equation of a conic in the form of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
(i) defines a parabola if $B^2-4AC=0$
(ii) defines an ellipse if $B^2-4AC\lt0$ and $A\ne C$
(iii) defines a circle if $B^2-4AC\lt0$ and $A= C$
(iv) defines a hyperbola if $B^2-4AC\gt0$
Here, we have
$A=3,B=-1,C=2$
Hence,
$B^2-4AC=(-1)^2-4(3)(2)=1-24=-23\lt0$ and $3\ne2$
Thus, it is an ellipse.
