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已知双曲线 ${\displaystyle C:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a>0,b>0)}$ 的离心率为 ${\displaystyle \sqrt{2}}$, 点 ${\displaystyle P_n}$ (${\displaystyle n=1,2,\cdots }$) 在 ${\displaystyle C}$ 上, 记 ${\displaystyle P_n}$ 的坐标为 ${\displaystyle (x_n,y_n)}$, ${\displaystyle x_n-y_n={\left(\frac{1}{2}\right)}^{n-3}}$, ${\displaystyle x_n>0}$, ${\displaystyle P_n}$ 关于 ${\displaystyle y}$ 轴的对称点为 ${\displaystyle Q_n}$. 已知 ${\displaystyle |P_1{Q}_1|=5}$.

1. 求 ${\displaystyle C}$ 的方程;
2. 证明: 直线 ${\displaystyle P_{n+1}{P}_{n+2}}$ 与直线 ${\displaystyle P_n{P}_{n+3}}$ 平行;
3. 设 ${\displaystyle \mathrm{\Delta }{P}_n{P}_{n+1}{P}_{n+2}}$ 的面积为 ${\displaystyle S_n}$, ${\displaystyle \mathrm{\Delta }{P}_n{P}_{n+1}{Q}_n}$ 的面积为 ${\displaystyle T_n}$. 当 ${\displaystyle \frac{T_n}{S_n}}$ 取得最小值时, 求 ${\displaystyle n}$.

1. 解 由题意, $$\begin{array}{r}\{\begin{array}{rl}& a=b,\\ & \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1,\\ & x_1-y_1=4,\\ & 2x_1=5,\end{array}\end{array}$$ 解得 ${\displaystyle C:x^2-y^2=4}$.
2. 证明 此时 $$4=x_n^2-y_n^2={\left(\frac{1}{2}\right)}^{n-3}(x_n+y_n),$$ 从而 ${\displaystyle x_n=2^{2-n}+2^{n-2}}$, ${\displaystyle y_n=2^{n-2}-2^{2-n}}$. 因此$$\begin{array}{rl}{k}_{P_{n+1}{P}_{n+2}}& =\frac{y_{n+2}-y_{n+1}}{x_{n+2}-x_{n+1}}=\frac{2^n+2^{1-n}}{2^n-2^{1-n}},\\ k_{P_n{P}_{n+3}}& =\frac{y_{n+3}-y_n}{x_{n+3}-x_n}=\frac{7(2^{n-1}+2^{-n})}{7(2^{n-1}-2^{-n})}=\frac{2^n+2^{1-n}}{2^n-2^{1-n}},\end{array}$$ 从而 ${\displaystyle P_{n+1}{P}_{n+2}//P_n{P}_{n+3}}$.
3. 解 由 2 可得, ${\displaystyle \mathrm{\forall }n\ge 2}$, ${\displaystyle P_n{P}_{n+1}//P_{n-1}{P}_{n+2}}$, 所以 $$S_n=S_{\mathrm{\Delta }{P}_n{P}_{n+1}{P}_{n+2}}=S_{\mathrm{\Delta }{P}_n{P}_{n+1}{P}_{n-1}}=S_{n-1},$$ 因此 ${\displaystyle S_n=S_{n-1}=\cdots =S_1}$ (也即 ${\displaystyle \{S_n\}}$ 是个常值数列), 因此 ${\displaystyle \frac{T_n}{S_n}}$ 的最小值其实就是 ${\displaystyle T_n}$ 的最小值. 而 $$T_n=\frac{1}{2}\cdot 2x_n(y_{n+1}-y_n)=\frac{1}{2}(3+2^{2n-3}+2^{4-2n}).$$ 记 ${\displaystyle f(x)=\frac{x}{2^3}+\frac{2^4}{x}}$, 有 ${\displaystyle f^{\prime }(x)=\frac{1}{2^3}-\frac{2^4}{x^2}}$, 有零点 ${\displaystyle 2^{\frac{7}{2}}}$, 因此 ${\displaystyle f(x)}$ 在 ${\displaystyle (0,2^{\frac{7}{2}})}$ 上减, ${\displaystyle (2^{\frac{7}{2}},+\mathrm{\infty })}$ 上增. 而 ${\displaystyle 2^{2n}{|}_{n=1}<2^{\frac{7}{2}}<2^{2n}{|}_{n=2}}$, 所以 ${\displaystyle T_n=\frac{1}{2}(3+f(2^{2n}))}$ 的最小值在 ${\displaystyle n=1}$ 或 ${\displaystyle 2}$ 处取得. 最后 ${\displaystyle T_1=\frac{15}{4}>T_2=3}$, 所以最小值在 ${\displaystyle n=2}$ 处取得.