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已知椭圆 ${\displaystyle C:{\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)}}$ 的左顶点为 ${\displaystyle A(-2,0)}$, 两个焦点与短轴一个顶点构成等腰直角三角形. 过点 ${\displaystyle P(1,0)}$ 且与 ${\displaystyle x}$ 轴不重合的直线 ${\displaystyle l}$ 与椭圆交于 ${\displaystyle M,N}$ 不同的两点.

1. 求椭圆 ${\displaystyle C}$ 的方程;
2. 当 ${\displaystyle AM}$ 与 ${\displaystyle MN}$ 垂直时，求 ${\displaystyle AM}$ 的长;
3. 过点 ${\displaystyle P}$ 且平行于 ${\displaystyle AM}$ 的直线交直线 ${\displaystyle x={\displaystyle \frac{5}{2}}}$ 于点 ${\displaystyle Q}$, 求证: 直线 ${\displaystyle NQ}$ 恒过定点.

1. 解 由题意 ${\displaystyle a=2,b=c=\sqrt{2}}$, 故 ${\displaystyle C:{\displaystyle \frac{x^2}{4}+\frac{y^2}{2}=1}}$.
2. 解 设 ${\displaystyle M(2\cos \alpha ,\sqrt{2}\sin \alpha )}$, 则由题意$$\begin{array}{r}0=\overrightarrow{MA}\cdot \overrightarrow{MP}=(-2-2\cos \alpha ,-\sqrt{2}\sin \alpha )\cdot (1-2\cos \alpha ,-\sqrt{2}\sin \alpha )=2{\cos }^2\alpha +2\cos \alpha ,\end{array}$$
   而 ${\displaystyle M\ne A}$，故解得 ${\displaystyle \cos \alpha =0,|\sin \alpha |=1}$，从而 ${\displaystyle AM=\sqrt{2^2+(\sqrt{2}{)}^2}=\sqrt{6}}$.
3. 证明 设 ${\displaystyle l_{MN}:x=ty+1,M(x_1,y_1),N(x_2,y_2)}$, 与椭圆联立得 $$\begin{array}{r}(t^2+2)y^2+2ty-3=0\Rightarrow y_1+y_2=-\frac{2t}{t^2+2},y_1{y}_2=-\frac{3}{t^2+2},\end{array}$$
   从而${\displaystyle 2t{y}_1{y}_2=3(y_1+y_2)}$. 另一方面，${\displaystyle l_{PQ}:{\displaystyle y=\frac{y_1}{x_1+2}(x-1)}}$，故${\displaystyle Q{\displaystyle (\frac{5}{2},\frac{3y_1}{2(x_1+2)})}}$.
   这样${\displaystyle NQ}$的横截距为$$\begin{array}{r}{x}_0=\frac{x_Q{y}_N-x_N{y}_Q}{y_N-y_Q}=\frac{\frac{5}{2}{y}_2-x_2\frac{3y_1}{2(x_1+2)}}{y_2-\frac{3y_1}{2(x_1+2)}}=\frac{2t{y}_1{y}_2+15y_2-3y_1}{2t{y}_1{y}_2+6y_2-3y_1}=2.\end{array}$$ 故 ${\displaystyle NQ}$ 恒过 ${\displaystyle (2,0)}$.