20. (10分)如图,在平面直角坐标系中,抛物线$y=x^{2}+bx+c$与$x$轴交于$A,B$两点,与$y$轴交于点$C$,已知$B(3,0)$,$C(0,-3)$,连接$BC$,点$P$是抛物线上的一个动点,点$N$是对称轴上的一个动点.
(1)求该抛物线的函数解析式;
(2)当$△ PAB$的面积为8时,求点$P$的坐标;
(3)若点$P$在直线$BC$的下方,当点$P$到直线$BC$的距离最大时,在抛物线上是否存在点$Q$,使得以点$P,C,N,Q$为顶点的四边形是平行四边形? 若存在,求出点$Q$的坐标;若不存在,请说明理由.

20.解:(1)$\because$ 抛物线$y=x^{2}+bx+c$经过点$B(3,0),C(0,-3)$,
$\therefore \{\begin{array}{l} c=-3,\\ 0=9+3b+c,\end{array} $ 解得$\{\begin{array}{l} b=-2,\\ c=-3.\end{array} $
$\therefore$ 抛物线的解析式为$y=x^{2}-2x-3$.
(2)$\because$ 抛物线$y=x^{2}-2x-3$与$x$轴交于$A,B$两点,
$\therefore 0=x^{2}-2x-3.\therefore x_{1}=-1,x_{2}=3$.
$\therefore$ 点$A(-1,0),AB=4$.
设点$P(p,p^{2}-2p-3)$,
$\because △PAB$的面积为 8,
$\therefore \frac {1}{2}×4×|p^{2}-2p-3|=8$.
$\therefore p^{2}-2p-3=4$或$p^{2}-2p-3=-4$.
$\therefore p_{1}=2\sqrt {2}+1,p_{2}=-2\sqrt {2}+1,p_{3}=1$.
$\therefore$ 点$P$的坐标为$(2\sqrt {2}+1,4)$或$(-2\sqrt {2}+1,4)$或$(1,-4)$.
(3)如图①,过点$P$作$PE⊥x$轴,交$BC$于$E$.

$\because$ 点$B(3,0),C(0,-3)$,
$\therefore$ 直线$BC$的解析式为$y=x-3$.
设点$P(a,a^{2}-2a-3)$,则点$E(a,a-3)$,
$\therefore PE=a-3-(a^{2}-2a-3)=-a^{2}+3a$.
$\therefore S_{△BCP}=\frac {1}{2}×(-a^{2}+3a)×3=-\frac {3}{2}(a-\frac {3}{2})^{2}+\frac {27}{8}$.
$\therefore$ 当$a=\frac {3}{2}$时,$S_{△BCP}$有最大值,即点$P$到直线$BC$的距离最大,
此时点$P(\frac {3}{2},-\frac {15}{4})$.
设点$N(1,n)$,点$Q(m,m^{2}-2m-3)$,
若$CP$为边,$CN$为边时,则$CQ$与$NP$互相平分,
$\therefore \frac {\frac {3}{2}+1}{2}=\frac {0+m}{2}$.
$\therefore m=\frac {5}{2}.\therefore$ 点$Q(\frac {5}{2},-\frac {7}{4})$.
若$CP$为边,$CQ$为边时,则$CN$与$PQ$互相平分,
$\therefore \frac {m+\frac {3}{2}}{2}=\frac {0+1}{2}$.
$\therefore m=-\frac {1}{2}.\therefore$ 点$Q(-\frac {1}{2},-\frac {7}{4})$.
若$CP$为对角线,则$CP$与$NQ$互相平分,
$\therefore \frac {0+\frac {3}{2}}{2}=\frac {1+m}{2}$.
$\therefore m=\frac {1}{2}.\therefore$ 点$Q(\frac {1}{2},-\frac {15}{4})$.
综上所述,点$Q$的坐标为$(\frac {5}{2},-\frac {7}{4})$或$(-\frac {1}{2},-\frac {7}{4})$或$(\frac {1}{2},-\frac {15}{4})$.