Adding to the direct answer
in terms of parabolas:
If you are reduced to ruler and compass
or just do not want to actually draw the two bounding parabolas,
you may want to transform the problem to a geometrically simpler one.
The method described below does this.
It can be extended to find slopes of asymptotes (if any) and of principal axes
when given five points of a conic section.
To simplify the description, I assume a projective geometry context.
Thus I may speak of points at infinity and the line at infinity.
Points at infinity occur as intersections of parallel lines and
can be interpreted as slopes of those lines.
A point is at infinity if and only if it lies on the line at infinity.
Relabel $p_1,\ldots,p_4$ to $A,B,C,D$. Any permutation is admissible.
(Some details will depend on the permutation, but the overall result won't.)
Let me also rename $p_5$ to $E$.
We will use $ABC$ as reference triangle below.
Relative to that reference triangle $ABC$, there is a transformation
of points called isogonal conjugation.
The inverse transformation is again isogonal conjugation.
However, isogonal conjugation of the vertices $A,B,C$ is undefined.
Therefore isogonal conjugation is a bijection only
for points outside the lines $\overline{AB}$, $\overline{BC}$, $\overline{CA}$.
In our use case, the resulting singularities can be removed by continuity.
Finding the isogonal conjugate of a point can be done using only ruler and
compass. Furthermore:
- Applying isogonal conjugation to the points of a line
yields a conic section through $A,B,C$ (a circumconic of $ABC$),
and every non-degenerate circumconic of $ABC$ can be obtained that way.
- Particularly, applying isogonal conjugation to all points at infinity
(i.e. the line at infinity) yields the circumcircle of $ABC$.
- Extension: The slope of a line (i.e. its point at infinity) determines
the slopes of the principal axes of the circumconic obtained by pointwise
isogonal conjugation of the line.
Given points $D,E$, we can construct their isogonal conjugates $D',E'$
and join them with a line $g = \overline{D'E'}$.
That line is the pointwise isogonal conjugate of the conic section through
$A,B,C,D,E$.
(You could now place a point $F'$ arbitrarily on $g$ and construct its
isogonal conjugate $F$ to obtain another point on the conic.)
Now consider the following cases:
If the line $g$ passes through one of the vertices of the reference triangle
$ABC$, e.g. $A$, then isogonal conjugation runs into a singularity. However,
algebraically, we can still argue that there exists a unique corresponding
circumconic, degenerated to a pair of lines, one of which passes through
that same vertex $A$, and the other matches the opposite side $\overline{BC}$.
Consequently, for a line through two vertices, e.g. $\overline{AB}$, the
corresponding conic consists of the other two extended sides $\overline{BC},
\overline{CA}$.
If $g$ intersects the circumcircle $U$ of $ABC$ in two distinct real points
$P_1',P_2'$, this means that the corresponding conic intersects
the line at infinity in two distinct points $P_1,P_2$ (at infinity).
A conic with two distinct points at infinity is a hyperbola
(possibly degenerated to a pair of nonparallel lines if case 1 applies),
and those points at infinity identify the slopes of its asymptotes.
Extension: $P_1,P_2$ can be obtained as isogonal conjugates of
$P_1',P_2'$. When breaking this down to more elementary steps,
each $P_i$ will arise as intersection of parallel lines.
Since you only need to know the associated slope,
you are done as soon as you have constructed one of those lines.
If $g$ touches the circumcircle $U$ of $ABC$ in one real point $P'$,
this means that the corresponding conic has a double point $P$ at infinity.
Such a conic is a parabola
(possibly degenerated to a pair of parallel lines if case 1 applies),
and its point at infinity gives the slope of the parabola's symmetry axis.
If $g$ does not intersect the circumcircle $U$ of $ABC$,
this means that the corresponding conic contains no point at infinity.
Such a conic is an ellipse.
Extension:
if $g$ is not the line at infinity,
the slopes of the corresponding circumconic's principal axes can be obtained by
determining the points $H_1',H_2'$ on the circumcircle $U$ with tangents
parallel to $g$ and transforming them back to points $H_1,H_2$ at infinity.
The solution to your problem is therefore to construct the isogonal conjugate
$D'$ of $D$ and its tangents to the circumcircle $U$ of $ABC$.
Then, when given another point $E$ of the conic, construct its isogonal
conjugate $E'$ and test whether $\overline{D'E'}$ intersects $U$.
The region for $E'$ where such intersection happens is bounded by the
tangents to $U$ through $D'$ and is tinted green in the figure below.
![Classification of a conic using isogonal conjugation "Classification of a conic using isogonal conjugation"](https://cdn.statically.io/img/i.sstatic.net/MJBeB.png)
- If $E'$ is inside the green region, the conic is a hyperbola,
possibly degenerated to a pair of nonparallel lines
(if $\overline{D'E'}$ intersects a vertex of $ABC$).
- If $E'$ lies on a bounding tangent, the conic is a parabola,
possibly degenerated to a pair of parallel lines
(if the point of tangency is a vertex of $ABC$).
- Otherwise the conic is an ellipse.
For those bounding tangents to exist, $D'$ must not be inside $U$.
Isogonal conjugation translates this requirement to $A,B,C,D$ being
vertices of a convex 4-gon, as given in the problem statement.