19 If there exist two points on the parabola $(y+1)^{2}=x+1$ that are symmetric about the line $y=x+a$, then the range of $a$ is
$19 a4\left(b^{2}+2 b\right)
$

i.e.
$
b>-\frac{9}{4} \text {. }
$

And (1) is $\frac{b-x_{1}+b-x_{2}}{2}=\frac{x_{1}+x_{2}}{2}+a$, so $b=\left(x_{1}+x_{2}\right)+a$. And $x_{1}+x_{2}=$ $2 b+3$, so $b=2 b+3+a, a=-b-3<\frac{9}{4}-3=-\frac{3}{4}$.