Let ABC be a triangle. Let A'B'C' be the orthic triangle of ABC. Draw a parabola focus at A', directrix B'C'. Then, the parabola touches BB' in T_b, CC' in T_c, AB in A_1 and AC in A_2.
Proof:
Consider the following configuration:
Now it is easy to see why BB', CC' are tangent to the parabola.
We know that the orthocenter of ABC is the incenter of the orthic triangle A'B'C'.
The tangency of AB and AC with the parabola follows directly from the Orthoptic property.
Addendum:
*Define B_1, B_2, C_1, C_2 as we define A_1, A_2.
Then, A_1, A_2, B_1, B_2, C_1, C_2 lie on a conic.
**Let the parabola cut the side BC in A_b, A_c. Define B_a, B_c, C_a, C_b cyclically. Then, these six points also lie on a conic.
Proof