The triangle $ABC$ is drawn on the plane. Using only a compass and a ruler, construct a focus and a directrix of the parabola $\Pi$ such that lines $AB$ and $AC$ are tangent to $\Pi$ at points $B$ and $C$ respectively.
My work. Let $X$ and $Y$ be the points of intersection of the directrix of the parabola $\Pi$ with the lines $AB$ and $AC$ respectively. Let $F$ be the focus of the parabola $\Pi$. I tried to apply the parabola property, which is that $\angle XFB=\angle YFC=\frac{\pi}{2}$, but didn't come up with anything else.
Consider C as the vertex of parabola. Now draw the diameter BC, find it's midpoint. The line LU connecting midpoint and A is parallel to the axis of symmetry of parabola, therefore it is perpendicular to it's directrix. A line L passing C, the vertex of parabola, and perpendicular to LU is the locus of points, including A, from which two tangents can be drawn on parabola such that AF is always perpendicular at one of these two tangent(F is supposed to be the focus). When A is coincident on C, CF is the axis of symmetry of parabola $L_1'$. So the line $L'_1$ passing C and parallel to LU is the axis of symmetry of parabola. For the focus Draw a perpendicular from A to AB, it intersects the axis of symmetry $L_1'$ at F which is the focus. The rest is easy; the distance form directrix and vertex v is equal to measure of CF, in this way you can also draw the dirctrix.