The normal to a curve at $P(x,y)$ meets the $x$ -axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is a:

The equation of the normal at $P(x,y)$ is: $Y-y=-\frac{dx}{dy}(X-x)$.The point $G$ is where this normal meets the $x$ -axis ($Y=0$):
$$-y=-\frac{dx}{dy}(X-x)⟹X=x+y\frac{dy}{dx}$$
So, $G=(x+y\frac{dy}{dx},0)$.Given that the distance of $G$ from the origin is twice the abscissa of $P$ ($x$):
$$|x+y\frac{dy}{dx}|=2|x|$$
Case 1: $x+y\frac{dy}{dx}=2x⟹y\frac{dy}{dx}=x⟹ydy=xdx$.Integrating: $\frac{y^2}{2}=\frac{x^2}{2}+C⟹y^2-x^2=2C$ (Hyperbola).Case 2: $x+y\frac{dy}{dx}=-2x⟹y\frac{dy}{dx}=-3x⟹ydy=-3xdx$.Integrating: $\frac{y^2}{2}=-\frac{3x^2}{2}+C⟹y^2+3x^2=2C$ (Ellipse).Thus, the curve can be an ellipse or a hyperbola.