8. 已知双曲线 $C:\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$（$a>0,b>0$）的左右焦点 $F_1$，$F_2$，左右顶点分别为 $A_1,A_2$，离心率为 $2$，点 $P$ 在 $C$ 上，若直线 $A_1P$，$A_2P$ 的斜率之和为 $\dfrac{6\sqrt{15}}{5}$，$\Delta PF_1F_2$ 的面积为 $\sqrt{15}$，则 $a=$
解：$\because e=\dfrac{c}{a}=2$，$a^{2}+b^{2}=c^{2}$
$\therefore c=2a$，$b=\sqrt{3}a$
设点 $P$ 在 $x$ 轴上方，且坐标为 $(x_0,y_0)$
依题可得
$\left\{ \begin{array}{l} \dfrac{x_0^{2}}{a^{2}}-\dfrac{y_0^{2}}{3a^{2}}=1 \\ \dfrac{y_0}{x_0-a}+\dfrac{y_0}{x_0+a}=\dfrac{6\sqrt{15}}{5} \\ \dfrac{1}{2}\cdot 4a\cdot y_0=\sqrt{15} \end{array} \right.$
$\therefore \left\{ \begin{array}{l} 3x_0^{2}-3a^{2}=y_0^{2} \\ \dfrac{2x_0y_0}{x_0^{2}-a^{2}}=\dfrac{6\sqrt{15}}{5} \\ y_0=\dfrac{\sqrt{15}}{2a} \end{array} \right.$
$\therefore x_0=\dfrac{3}{2a}$
$\therefore 3\cdot\left( \dfrac{3}{2a} \right)^{2}-3a^{2}=\left( \dfrac{\sqrt{15}}{2a} \right)^{2}$
$\therefore a=1$
16. 若直线 $y=kx+b$ 是曲线 $y=\dfrac{\ln x}{x}$ 的切线，也是曲线 $y=\dfrac{2}{x}$ 的切线，则 $k=$
解：设直线 $y=kx+b$ 与曲线 $y=\dfrac{\ln x}{x}$ 的切点为 $\left( x_1,\dfrac{\ln x_1}{x_1} \right)$
$\because y'=\dfrac{1-\ln x}{x^{2}}$
$\therefore$ 切线方程为 $y-\dfrac{\ln x_1}{x_1}=\dfrac{1-\ln x_1}{x_1^{2}}(x-x_1)$
设直线 $y=kx+b$ 与曲线 $y=\dfrac{2}{x}$ 的切点为 $\left( x_2,\dfrac{2}{x_2} \right)$
$\because y'=-\dfrac{2}{x^{2}}$
$\therefore$ 切线方程为 $y-\dfrac{2}{x_2}=-\dfrac{2}{x_2^{2}}(x-x_2)$
$\therefore \left\{ \begin{array}{l} \dfrac{1-\ln x_1}{x_1^{2}}=-\dfrac{2}{x_2^{2}} \\ \dfrac{\ln x_1}{x_1}-\dfrac{1-\ln x_1}{x_1}=\dfrac{2}{x_2}+\dfrac{2}{x_2} \end{array} \right.$
$\therefore \left\{ \begin{array}{l} \dfrac{\ln x_1-1}{2}=\dfrac{x_1^{2}}{x_2^{2}} \\ \dfrac{2\ln x_1-1}{4}=\dfrac{x_1}{x_2} \end{array} \right.$
$\therefore \dfrac{\ln x_1-1}{2}=\left( \dfrac{2\ln x_1-1}{4} \right)^{2}$
$\therefore \ln x_1=\dfrac{3}{2}$，$x_1=e^{\frac{3}{2}}$
$\therefore k=\dfrac{1-\ln x_1}{x_1^{2}}=-\dfrac{1}{2}e^{-3}$