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 $A=1,B=-2,C=3$, hence,
$B^2-4AC=(-2)^4-4(1)(3)=4-12=-8\lt0$ and $1\ne3$
Thus, it is an ellipse.
