所给椭圆即 $\frac{(x-3)^2}{5}+y^2=1$．其左焦点为 $F_1(1,0)$．设 $A$ 的坐标为 $(x_1,y_1)$，$B$ 的坐标为 $(x_2,y_2)$，则由 $F_1A\perp F_1B$，知$$\overrightarrow{F_1A}\cdot \overrightarrow{F_1B}=(x_1-1)(x_2-1)+y_1y_2=0.$$即$$y_1y_2=(x_1+x_2)-x_1x_2-1.~~~~~ ① $$将 $y=kx$ 代入椭圆方程，消去 $y$，整理得$$(5k^2+1)x^2-6x+4=0.$$故$$x_1+x_2=\frac{6}{5k^2+1}, x_1x_2=\frac{4}{5k^2+1}.~~~~~$$由 $A,B$ 在直线 $l$ 上，得$$y_1y_2=k^2x_1x_2=\frac{4k^2}{5k^2+1}.$$由 $A,B$ 在直线 $l$ 上，得$$y_1y_2=k^2x_1x_2=\frac{4k^2}{5k^2+1}.$$将式 ②，③ 代入式 ①，得$$\frac{4k^2}{5k^2+1}=\frac{6}{5k^2+1}-\frac{4}{5k^2+1}-1,$$解得 $k^2=\frac{1}{9}$，即 $k=\pm \frac{1}{3}$．