The center of the circular luminous object is $C$, and its radius is $r$. The center of the thin lens is $O$, and its focal length is $f$. The object is in the same plane as $C$, and $OC$ is perpendicular to the thin lens, and $\left\|OC\right\|=2f$.
Set up Cartesian coordinate with $O$ being the origin and $CO$ being the $x$-axis. Then, $C\left(-2f.0\right)$. The luminous object is described by the parametric equations $\begin{cases} x=-2f+r\cos t,\\y=r\sin t. \end{cases}$ Pick point $P\left(-2f+r\cos t,r\sin t\right)$ on the object. According to the formula for imaging of thin lenses $\begin{cases} -\frac1x+\frac1{x'}=\frac1f,\\ \frac x{x'}=\frac y{y'}, \end{cases}$ the point $P$ is transformed to $P'\left(f\left(1+\frac f{f-r\cos t}\right),-f\frac{r\sin t}{f-r\cos t}\right)$. Therefore, we can have the parametric equations of the image: $\begin{cases} x=f\left(1+\frac f{f-r\cos t}\right),\\y=-f\frac{r\sin t}{f-r\cos t}. \end{cases}$ Cancel $t$, and we have $y^2=\left(\frac rf\right)^2\left(x-f\right)^2-\left(x-2f\right)^2,$ which means that the image is a conic section with the focus being $\left(2f,0\right)$, the directrix being line $x=f$, and the eccentricity being $\frac rf$.
Alternatively, let $\rho\coloneqq\sqrt{\left(x-2f\right)^2+y^2}$, and we have $\rho=\frac r{1-\frac rf\cos t},$ and we may have the same conclusion through this equation.