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已知椭圆 ${\displaystyle E:{\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)}}$ 过点 ${\displaystyle A(3,0)}$, 焦距为 ${\displaystyle 2\sqrt{5}}$.

1. 求椭圆 ${\displaystyle E}$ 的方程, 并求其短轴长;
2. 过点 ${\displaystyle P(1,0)}$ 且不与 ${\displaystyle x}$ 轴重合的直线 ${\displaystyle l}$ 交椭圆 ${\displaystyle E}$ 于两点 ${\displaystyle C,D}$, 连 ${\displaystyle CO}$ 并延长交椭圆 ${\displaystyle E}$ 于点 ${\displaystyle M}$, 直线 ${\displaystyle AM}$ 与 ${\displaystyle l}$ 交于点 ${\displaystyle N}$, ${\displaystyle Q}$ 为 ${\displaystyle OD}$ 的中点，其中 ${\displaystyle O}$ 为原点. 设直线 ${\displaystyle NQ}$ 的斜率为 ${\displaystyle k}$, 求 ${\displaystyle k}$ 的最大值.

1. 由题意得 ${\displaystyle a=3,c=\sqrt{5}}$, 故 ${\displaystyle b=2,E:{\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1}}$, 短轴长为 ${\displaystyle 4}$.
2. 设 ${\displaystyle C(x_1,y_1),D(x_2,y_2)}$.设 ${\displaystyle l:x=ty+1}$，与椭圆联立得$$\begin{array}{r}(4t^2+9)y^2+8ty-32=0\Rightarrow y_1+y_2=-\frac{8t}{4t^2+9}.\end{array}$$
   则 ${\displaystyle M(-x_1,-y_1),l_{AM}:{\displaystyle y=\frac{y_1}{x_1+3}(x-3)}}$, 与 ${\displaystyle l}$ 联立解得 ${\displaystyle N{\displaystyle (-\frac{t{y}_1}{2}+1,-\frac{y_1}{2})}}$.又 ${\displaystyle Q{\displaystyle (\frac{1}{2}{x}_2,\frac{1}{2}{y}_2)=(\frac{t{y}_2+1}{2},\frac{y_2}{2})}}$,故 $$\begin{array}{r}k=\frac{\frac{y_2}{2}+\frac{y_1}{2}}{\frac{t{y}_2+1}{2}+\frac{t{y}_1}{2}-1}=\frac{y_1+y_2}{t(y_1+t_2)-1}=\frac{8t}{12t^2+9}\le \frac{8|t|}{12|t{|}^2+9}\le \frac{2\sqrt{3}}{9}.\end{array}$$