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已知双曲线 ${\displaystyle C}$ 的虚轴长为 ${\displaystyle 2}$，其中一条渐近线方程为 ${\displaystyle y={\displaystyle \frac{1}{2}x}}$, 且 ${\displaystyle M,N}$ 分别是双曲线的左、右顶点.

1. 求双曲线 ${\displaystyle C}$ 的方程;
2. 设过点 ${\displaystyle G(4,0)}$ 的动直线 ${\displaystyle l}$ 交双曲线 ${\displaystyle C}$ 右支于 ${\displaystyle A,B}$ 两点，若直线 ${\displaystyle AM,BN}$ 的斜率分别为 ${\displaystyle k_1,k_2}$.
   1. 试探究 ${\displaystyle k_1}$ 与 ${\displaystyle k_2}$ 的比值 ${\displaystyle {\displaystyle \frac{k_1}{k_2}}}$ 是否为定值. 若是定值，求出这个定值; 若不是定值，请说明理由;
   2. 设 ${\displaystyle \mathrm{\angle }ANG=\alpha ,\mathrm{\angle }BNG=\beta ,0<\beta <{\displaystyle \frac{\pi }{2}}}$, 若 ${\displaystyle {\displaystyle \tan \theta =\frac{1}{7},\alpha =\beta -\theta (0<\theta <\frac{\pi }{2})}}$, 求 ${\displaystyle \mathrm{\Delta }BGN}$ 的面积.

1. 双曲线 ${\displaystyle C}$ 的方程是 ${\displaystyle {\displaystyle \frac{x^2}{4}-y^2=1}}$.
2. 1. 由于 ${\displaystyle l}$ 与 ${\displaystyle C}$ 在右支交于两点，故 ${\displaystyle l}$ 的斜率不会为零. 又 ${\displaystyle l}$ 恒过 ${\displaystyle G(4,0)}$，故设 ${\displaystyle l:x=ty+4}$. 与 ${\displaystyle C}$ 联立得 $$\begin{array}{}\text{(*)}& (t^2-4)y^2+8ty+12=0\end{array}$$
      设 ${\displaystyle A(x_1,y_1),B(x_2,y_2)}$.由韦达定理得 $$\begin{array}{r}{y}_1{y}_2=\frac{12}{t^2-4},y_1+y_2=-\frac{8t}{t^2-4}\Rightarrow t{y}_1{y}_2=-\frac{3}{2}(y_1+y_2).\end{array}$$
      代入${\displaystyle {\displaystyle k_1=\frac{y_1}{x_1+2},k_2=\frac{y_2}{x_2-2}}}$,得$$\begin{array}{r}\frac{k_1}{k_2}=\frac{y_1}{y_2}\cdot \frac{x_2-2}{x_1+2}=\frac{y_1}{y_2}\cdot \frac{t{y}_2+2}{t{y}_1+6}=\frac{t{y}_1{y}_2+2y_1}{t{y}_1{y}_2+6y_2}=\frac{-{\displaystyle \frac{3}{2}}(y_1+y_2)+2y_1}{-{\displaystyle \frac{3}{2}}(y_1+y_2)+6y_2}=-\frac{1}{3}.\end{array}$$
   2. 由题意，${\displaystyle k_{BN}=-{\displaystyle \frac{y_2}{x_2-2},k_{AN}=\frac{y_1}{x_1-2}}}$.由于 ${\displaystyle k_{AN},k_{BN}}$ 异号，不妨设${\displaystyle {\displaystyle \tan \beta =-\frac{y_2}{x_2-2},\tan \alpha =\frac{y_1}{x_1-2}}}$ (最后求解出的 ${\displaystyle |t|}$ 是相同的.) 因此 $$\begin{array}{rl}\frac{1}{7}=\tan (\beta -\alpha )& =\frac{\tan \beta -\tan \alpha }{1+\tan \alpha \tan \beta }=-\frac{2t{y}_1{y}_2+2(y_1+y_2)}{(t^2-1)y_1{y}_2+2t(y_1+y_2)+4}=\frac{2t}{7},\end{array}$$
      从而 ${\displaystyle {\displaystyle t=\frac{1}{2}}}$.此时 ${\displaystyle k_{AB}>0}$,为使 ${\displaystyle \beta =\alpha +\theta >\alpha }$,则 ${\displaystyle y_B<0<y_A}$,因此将 ${\displaystyle t}$ 回代入 ${\displaystyle (\ast )}$ 解得 ${\displaystyle y_B={\displaystyle -\frac{4}{3}}}$.从而 ${\displaystyle {\displaystyle S_{\mathrm{\Delta }BGN}=\frac{4}{3}}}$.