Conic Associated to Three Parabolas and a Triangle

Let ABC be a triangle.
Let B and side AC be the focus and directrix of a parabola.
Let C and side AB be the focus and directrix of a parabola.
Let A and side BC be the focus and directrix of a parabola.
Let the black parabola intersect sides AB, BC in D, I respectively.
Let the blue parabola intersect sides AB, AC in E, F respectively.
Let the red parabola intersect sides AC, BC in G, H respectively.

Then, E, I, D, F, G, H lie on a conic.

Proof:

Proof:
AA', BB' CC' be the three altitudes

From the definition of parabola we have:
BD = DL; AF = FN; CH = HJ; BI = IM; CG = GK; AE = EO

By similarity we have:
BD = BB'*AD/AB
BE = AE*AB/AA'
AF = AA'*CF/AC
AG = CG*AC/CC'
CH = CC'*BH/BC
CI = BI*BC/BB'

From the Carnot's Theorem for conics:
BD*BE*AF*AG*CH*CI =
(BB'*AD/AB)(AE*AB/AA')(AA'*CF/AC)(CG*AC/CC')(CC'*BH/BC)(BI*BC/BB')= AD*AE*CF*CG*BH*BI.

So done.

Generalization