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1. 如图，在平面直角坐标系中，抛物线$y = - x^{2} + 2x + 3$与$x$轴交于点$A$，$B$，与$y$轴交于点$C$，作直线 BC.$P$是线段 BC 上方的抛物线上一动点，过点$P$作$PQ \perp BC$，垂足为$Q$，则线段 PQ 的最大值为
$\frac{9\sqrt{2}}{8}$
，此时点$P$的坐标为$\left(\frac{3}{2},\frac{15}{4}\right)$
.
答案：
1.$\frac{9\sqrt{2}}{8}$ $\left(\frac{3}{2},\frac{15}{4}\right)$ 点拨:如答图，过点P作PN⊥AB于点N，交BC于点M.令$-x^{2}+2x+3=0$，解得$x_{1}=-1$，$x_{2}=3$，$\therefore A(-1,0)$，$B(3,0)$.令$x=0$，则$y=3$，$\therefore C(0,3)$.
$\because B(3,0)$，$C(0,3)$，$\therefore$直线BC的函数表达式为$y=-x+3$.
$\because OB=OC$，$\angle BOC=90^{\circ}$，$\therefore\angle CBO=45^{\circ}$. $\because\angle MNB=90^{\circ}$，$\therefore\angle PMQ=\angle NMB=45^{\circ}$. $\because PQ\perp BC$，$\therefore\triangle PQM$是等腰直角三角形，$\therefore PM=\sqrt{2}PQ$，$\therefore PM$的值最大时，$PQ$的值最大.
设$P(m,-m^{2}+2m+3)$，则$M(m,-m+3)$，$\therefore PM=-m^{2}+2m+3-(-m+3)=-m^{2}+3m$.
$\because -1＜0$，$\therefore$当$m=\frac{3}{2}$时，$PM$的值最大，$PM$的最大值为$-\frac{9}{4}+\frac{9}{2}=\frac{9}{4}$，
$\therefore PQ$的最大值为$\frac{\sqrt{2}}{2}PM=\frac{9\sqrt{2}}{8}$，此时点P的坐标为$\left(\frac{3}{2},\frac{15}{4}\right)$.
1.$\frac{9\sqrt{2}}{8}$ $\left(\frac{3}{2},\frac{15}{4}\right)$ 点拨:如答图，过点P作PN⊥AB于点N，交BC于点M.令$-x^{2}+2x+3=0$，解得$x_{1}=-1$，$x_{2}=3$，$\therefore A(-1,0)$，$B(3,0)$.令$x=0$，则$y=3$，$\therefore C(0,3)$.
$\because B(3,0)$，$C(0,3)$，$\therefore$直线BC的函数表达式为$y=-x+3$.
$\because OB=OC$，$\angle BOC=90^{\circ}$，$\therefore\angle CBO=45^{\circ}$. $\because\angle MNB=90^{\circ}$，$\therefore\angle PMQ=\angle NMB=45^{\circ}$. $\because PQ\perp BC$，$\therefore\triangle PQM$是等腰直角三角形，$\therefore PM=\sqrt{2}PQ$，$\therefore PM$的值最大时，$PQ$的值最大.
设$P(m,-m^{2}+2m+3)$，则$M(m,-m+3)$，$\therefore PM=-m^{2}+2m+3-(-m+3)=-m^{2}+3m$.
$\because -1＜0$，$\therefore$当$m=\frac{3}{2}$时，$PM$的值最大，$PM$的最大值为$-\frac{9}{4}+\frac{9}{2}=\frac{9}{4}$，
$\therefore PQ$的最大值为$\frac{\sqrt{2}}{2}PM=\frac{9\sqrt{2}}{8}$，此时点P的坐标为$\left(\frac{3}{2},\frac{15}{4}\right)$.
2. 如图是直线$l:y_{1} = x + 4$和抛物线$y_{2} = - x^{2} + 2x$在同一平面直角坐标系中，$P$为抛物线上一动点，则点$P$到直线$l$的距离的最小值为
$\frac{15\sqrt{2}}{8}$
.
答案：
2.$\frac{15\sqrt{2}}{8}$ 点拨:如答图，过点P作$PH\perp y$轴于点H，作PQ//y轴，交直线$y = x + 4$于点Q.
设$P(t,-t^{2}+2t)$，$Q(t,t+4)$，则$PQ=t+4-(-t^{2}+2t)=t^{2}-t+4$.
$\because y = x + 4$，$\therefore\angle PQH=45^{\circ}$，
$\therefore PH=\frac{\sqrt{2}}{2}PQ=\frac{\sqrt{2}}{2}(t^{2}-t+4)=\frac{\sqrt{2}}{2}\left(t-\frac{1}{2}\right)^{2}+\frac{15\sqrt{2}}{8}$.
$\because\frac{\sqrt{2}}{2}＞0$，$\therefore$当$t=\frac{1}{2}$时，$PH$有最小值，最小值为$\frac{15\sqrt{2}}{8}$.
2.$\frac{15\sqrt{2}}{8}$ 点拨:如答图，过点P作$PH\perp y$轴于点H，作PQ//y轴，交直线$y = x + 4$于点Q.
设$P(t,-t^{2}+2t)$，$Q(t,t+4)$，则$PQ=t+4-(-t^{2}+2t)=t^{2}-t+4$.
$\because y = x + 4$，$\therefore\angle PQH=45^{\circ}$，
$\therefore PH=\frac{\sqrt{2}}{2}PQ=\frac{\sqrt{2}}{2}(t^{2}-t+4)=\frac{\sqrt{2}}{2}\left(t-\frac{1}{2}\right)^{2}+\frac{15\sqrt{2}}{8}$.
$\because\frac{\sqrt{2}}{2}＞0$，$\therefore$当$t=\frac{1}{2}$时，$PH$有最小值，最小值为$\frac{15\sqrt{2}}{8}$.
3. 如图，抛物线$y = - \frac{1}{2}x^{2} + \frac{3}{2}x + 2$与$x$轴交于点$A$，$B$，与$y$轴交于点$C$，顶点为$M$，连接 BC. 若$D$是直线 BC 上方抛物线上一动点，连接 OD 交 BC 于点$E$，当$\frac{DE}{OE}$的值最大时，求点 D 的坐标.
答案：
3.解:如答图，过点D作$DH// y$轴，交BC于点H.
设$D\left(m,-\frac{1}{2}m^{2}+\frac{3}{2}m+2\right)$.在$y=-\frac{1}{2}x^{2}+\frac{3}{2}x+2$中，令$x=0$，则$y=2$，$\therefore C(0,2)$.令$y=0$，则$-\frac{1}{2}x^{2}+\frac{3}{2}x+2=0$，解得$x_{1}=-1$，$x_{2}=4$，$\therefore A(-1,0)$，$B(4,0)$.
由点$B(4,0)$，$C(0,2)$，知直线BC的函数表达式为$y=-\frac{1}{2}x+2$，$\therefore H\left(m,-\frac{1}{2}m+2\right)$，
$\therefore DH=-\frac{1}{2}m^{2}+\frac{3}{2}m+2-(-\frac{1}{2}m+2)=-\frac{1}{2}m^{2}+2m$. $\because DH// y$轴，$\therefore\triangle OCE\backsim\triangle DHE$，
$\therefore\frac{DE}{OE}=\frac{DH}{CO}=\frac{-\frac{1}{2}m^{2}+2m}{2}=-\frac{1}{4}(m - 2)^{2}+1$.
$\because-\frac{1}{4}＜0$，$\therefore$当$m = 2$时，$\frac{DE}{OE}$的值最大，此时点D的坐标为$(2,3)$.
3.解:如答图，过点D作$DH// y$轴，交BC于点H.
设$D\left(m,-\frac{1}{2}m^{2}+\frac{3}{2}m+2\right)$.在$y=-\frac{1}{2}x^{2}+\frac{3}{2}x+2$中，令$x=0$，则$y=2$，$\therefore C(0,2)$.令$y=0$，则$-\frac{1}{2}x^{2}+\frac{3}{2}x+2=0$，解得$x_{1}=-1$，$x_{2}=4$，$\therefore A(-1,0)$，$B(4,0)$.
由点$B(4,0)$，$C(0,2)$，知直线BC的函数表达式为$y=-\frac{1}{2}x+2$，$\therefore H\left(m,-\frac{1}{2}m+2\right)$，
$\therefore DH=-\frac{1}{2}m^{2}+\frac{3}{2}m+2-(-\frac{1}{2}m+2)=-\frac{1}{2}m^{2}+2m$. $\because DH// y$轴，$\therefore\triangle OCE\backsim\triangle DHE$，
$\therefore\frac{DE}{OE}=\frac{DH}{CO}=\frac{-\frac{1}{2}m^{2}+2m}{2}=-\frac{1}{4}(m - 2)^{2}+1$.
$\because-\frac{1}{4}＜0$，$\therefore$当$m = 2$时，$\frac{DE}{OE}$的值最大，此时点D的坐标为$(2,3)$.
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