Family of parabolas with given directrix and incident point What is the most elegant description of the family of parabolas with a given directrix $d$ and through a given point $P$? And what is the description of the union of all these parabolas in the real plane?
This question was inspired by me initially misreading another question. The goal of “elegant” is deliberately vague to allow for interesting answers from others even though I intend to post my own answer. Don't let my background dictate your approach.
 A: A parabola is the set of all points having the same distance to the directrix $d$ and the focus $F$. So by knowing $d$ and a point $P$ on the parabola, you know that distance for $P$. Therefore $F$ has to be a point at the same distance from $P$, and not on $d$. So $F$ is restricted to a circle except for one point, and the whole setup can be described by one real parameter.
The whole configuration is invariant under similarity transformations. So without loss of generality you can pick $P=(0,0)$ and $d:x=1$ so that the circle on which $F$ must lie is the unit circle. Using the tangent half-angle substitution and homogeneous coordinates, you can describe the focus as
$$F=[t^2-1:2t:t^2+1]\qquad\text{with}\quad t\in\mathbb R$$
Now when does a point $Q=(x,y)$ lie on that parabola?
\begin{align*}
\lvert Q,d\rvert^2 &= \lvert Q,F\rvert^2 \\
(x-1)^2+y^2 &=
\left(x-\frac{t^2-1}{t^2+1}\right)^2+\left(y-\frac{2t}{t^2+1}\right)^2 \\
x &= ty-\tfrac14(t^2+1)y^2
\end{align*}
Or to write this conic as a matrix:
$$(x,y,1)\cdot\begin{pmatrix}0&0&2\\0&t^2+1&-2t\\2&-2t&0\end{pmatrix}
\cdot\begin{pmatrix}x\\y\\1\end{pmatrix}=0$$
At this point I had hoped to write the result as some linear combination of simple conics, but apparently it isn't as easy as that. So there is room here for improvement by other answers.
To find a description of the union of all these points, start by taking $Q=(x,y)$ then find a $t$ to satisfy the equation. Since the equation is quadratic in $t$ you get zero, one or two possible solutions. The boundary is exactly the limit between zero and two solutions. Write the conic as a polynomial in $t$ namely
$$(y^2)t^2-(4y)t+(y^2+4x)=0$$
then compute its discriminant with respect to $t$, which (according to Sage) is
$$-4y^2(y^2+4x-4)$$
This polynomial is zero at the boundary. The right parenthesis there is the outer boundary of the union, again a parabola. The $y^2$ term refers to the degenerate situation where the line $PQ$ is orthogonal to $d$ and the parabola becomes a double line. That situation needs to be excluded if the question statement is interpreted to refer to non-degenerate parabolas only. However, one point with $y=0$ is part of the union, namely $P$ which lies on every parabola. So the union of all non-degenerate parabolas is given as
$$\{(x,y)\mid 0<y^2\le4(1-x)\}\cup\{(0,0)\}$$
