This actually works for any 2 degree plane curve (aka conics) We show that if $E\in\mathcal{C}$ then $F\in\mathcal{C}$, the reverse direction follows reversely. Let $\overline{BF}\cap S_2=\{P\}$ and $\overline{AE}\cap S_2=\{Q\}$. Then by Reim's $\overline{PQ}\|\overline{CD}$. So $\angle QEC=\angle PED=\angle PFD$ $(\bigstar)$. Let $\overline{QE}\cap\overline{CF}=\{X\}$ and $\overline{DE}\cap\overline{PF}=\{Y\}$. Then from $(\bigstar)$ we get $EFXY$ cyclic $\implies XY\|AB\|CD$ by Reims $\implies AB\cap CD\cap XY=\infty$. So by Converse of Pascal on $ABFCDE$ we get that $F\in\mathcal C$. So in this case $\{A,B,C,D,E\}\in\mathcal {P}$(Parabola). So $F\in\mathcal{P}$. $\blacksquare$

Lemma. Let points $A(a,ka^2),B(b,kb^2),C(c,kc^2)$ are points of intersection of parabola $y=kx^2$ with circle then center has coordinates $(-\frac{k^2(a+b)(b+c)(c+a)}{2},\frac{k^2(a^2+b^2+c^2+ab+bc+ca)+1}{2k})$ We have $(x-a)^2+(y-ka^2)=(x-b)^2+(y-kb^2)=(x-c)^2+(y-kc^2)$ Then $(b-a)(2x-a-b)=(ka^2-kb^2)(2y-ka^2-kb^2)$ $2y= \frac{-2x}{k(a+b)}+\frac{1}{k}+k(a^2+b^2)$ and $2y=\frac{-2x}{k(a+c)}+\frac{1}{k}+k(a^2+c^2)$ $\frac{-2x}{k(a+b)}+\frac{1}{k}+k(a^2+b^2)=\frac{-2x}{k(a+c)}+\frac{1}{k}+k(a^2+c^2)$ $x=-\frac{k^2(c+b)(a+b)(a+c)}{2}$ $y=\frac{1}{2} (\frac{-2x}{k(a+b)}+\frac{1}{k}+k(a^2+b^2))=\frac{1}{2} ( k(a+c)(b+c)+\frac{1}{k}+k(a^2+b^2))= \frac{1}{2} ( k (a^2+b^2+c^2+ab+bc+ca)+\frac{1}{k})$ End of proof Lemma 2. If points $A(a,ka^2),B(b,kb^2),C(c,kc^2),D(d,kd^2)$ are points of intersection of circle with parabola $y=kx^2$ then $a+b+c+d=0$ As circumcenter of $ABC$ and $ABD$ is same then $ (a+b)(b+c)(c+a)=(a+b)(b+d)(d+a) \to c^2+(a+b)c=d^2+(a+c)d \to a+b+c+d=0$ End of proof. We can replace parabola $y=kx^2+px+q$ with parabola $y=kx^2$ Let points $A,B,C,D,E$ has coordinates $(a,ka^2),(b,kb^2),(c,kc^2),(d,kd^2),(e,ke^2)$ $AB\parallel CD \to a+b=c+d$ Let point $T$ is other point of intersection of circumcircle of $ABE$ and parabola. Such point exists because $(x-x_0)^2+(kx^2-y_0)^2=r^2$ has $4$ solutions Then $T=(-(a+b+e),k(a+b+e)^2)$ But we have $a+b=c+d$ so $T=(-(c+d+e),(c+d+e)^2)$ is point of intersection of circumcircle of $CDE$ with parabola. So point $T$ is point of intersection of circumcircles and lies on parabola

Dear Mathlinkers, it can be seen as a generalization of the Reim's theorem...when the two lines become a parabola... Sincerely Jean-Louis