How to determine number of parabolas tangent to three given lines, or tangent to two given lines at specific points I recently came across a question which prompted me to think how we could determine the uniqueness of a parabola

How many parabolas would satisfy the following conditions:

*

*Tangent to $3$ given (evidently non-concurrent) lines

*Tangent to $2$ given (evidently non-parallel) lines at specific points on the lines; eg, tangent to $y=0$ at $(12, 0)$ and to $y+x=5$ at the point $(4, 1)$

I heard somewhere that a parabola requires $4$ degrees of freedom, I'd also appreciate an explanation in context to this statement.
In the latter case, how would we find the equation of the parabola, if a unique one does exist?
 A: Not really an answer, but I would like to show a very simple geometric construction to find the parabola tangent to four given lines, thus also providing a proof of its uniqueness.
The construction relies on a beautiful property of parabolas:

The focus of a parabola lies on the circumcircle of the triangle
formed by any three tangents to the parabola (see Appendix for a
geometric proof).

Let then $a$, $b$, $c$, $c$ be four given lines (no two of them parallel). The focus $F$ of the parabola tangent to them must lie (by the above property) on the circumcircle of the triangle formed by lines $abc$, and at the same time it must lie on the circumcircle of the triangle formed by lines $abd$. Those two circles have in common the intersection point of lines $ab$, which cannot be the focus. Hence the focus of the parabola is uniquely determined as the other intersection of the circumcircles.
To complete the construction we can then use another well-known property:

The perpendicular projection of the focus of a parabola onto any
tangent, lies on the line passing through the vertex and perpendicular
to the axis of the parabola.

If $A$ and $B$ are the projections of the focus on lines $a$ and $b$, the line through the focus perpendicular to line $AB$ is then the axis, and the intersection of the lines is the vertex of the parabola.
APPENDIX.
Proof of the property given at the beginning.
Let three tangents to a parabola be given, with tangency points $A$, $B$, $C$ and intersecting at $P$, $Q$, $R$ (see figure below). Let $F$ be the focus of the parabola, and $D$, $E$ the intersections of the axis of the parabola with lines $PQ$ and $QR$.
It is well known that $FA=FD$, which entails
$\angle FAQ=\angle EDQ$.
In addition, we know that (see here for a proof):

the exterior angle between any two tangents is equal to the angle
which either segment of tangent subtends at the focus,

which in our case implies $\angle AFQ=\angle DQE$.
It follows that in triangles $AFQ$, $DEQ$ we also have
$\angle FQA=\angle DEQ$.
But with an analogous reasoning we can prove that
$\angle FRB=\angle FER=\angle DEQ$. It follows that
$\angle FRP=\angle FQP$, and points $FPQR$ are concyclic as it was to be proved.

