Two parabolas have the same focus, namely the point $(3,-28).$  Their directrices are the $x$-axis and the $y$-axis, respectively.  Compute the slope of their common chord.
Let $P$ and $Q$ be the the intersection points of the two parabolas.  Then by definition of the parabola, the distance from $P$ to their common focus $F$ is equal to the distance from $P$ to the $x$-axis.  Also, the distance between $P$ to $F$ is equal to $P$ to the $y$-axis.  This means $P$ is equidistant to both the $x$-axis and $y$-axis, so $P$ must lie on the line $y = -x.$

[asy]
unitsize(0.15 cm);

pair F, P, Q;

real parab (real x) {
  return(-(x^2 - 6*x + 793)/56);
}

real upperparab(real x) {
  return(sqrt(3)*sqrt(2*x - 3) - 28);
}

real lowerparab(real x) {
  return(-sqrt(3)*sqrt(2*x - 3) - 28);
}

F = (3,-28);
P = (18.0385,-18.0385);
Q = (43.9615,-43.9615);

draw((-10,0)--(50,0));
draw((0,-50)--(0,10));
draw(graph(parab,-10,47),red);
draw(graph(upperparab,3/2,50),blue);
draw(graph(lowerparab,3/2,50),blue);
draw(F--P--(P.x,0));
draw(P--(0,P.y));
draw(F--Q--(Q.x,0));
draw(Q--(0,Q.y));
draw((-10,10)--(50,-50),dashed);

label("$F$", F, NW, UnFill);
label("$P$", P, NE, UnFill);
label("$Q$", Q, NE, UnFill);

dot(F);
[/asy]

By the same argument, $Q$ also lies on the line $y = -x.$  Therefore, the slope of $\overline{PQ}$ is $\boxed{-1}.$