10. 已知抛物线 $y^{2}=-x$ 与直线 $l: y=k(x+1)$ 相交于 $A, B$ 两点.

(1)求证: $O A \perp O B$;

(2)当 $\triangle O A B$ 的面积等于 $\sqrt{10}$ 时, 求 $k$ 的值.
：解： (1)联立 $\left\{\begin{array}{l}y^{2}=-x, \\ y=k(x+1),\end{array}\right.$

消去 $x$, 得 $k y^{2}+y-k=0$.

设 $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,

则 $y_{1}+y_{2}=-\frac{1}{k}, y_{1} \cdot y_{2}=-1$.

$\because y_{1}^{2}=-x_{1}, \quad y_{2}^{2}=-x_{2}, \quad \therefore\left(y_{1} \cdot y_{2}\right)^{2}=x_{1} \cdot x_{2}$.

$\therefore x_{1} \cdot x_{2}=1 . \therefore x_{1} x_{2}+y_{1} y_{2}=0$,

即 $\overrightarrow{O A} \cdot \overrightarrow{O B}=0 . \therefore O A \perp O B$.

(2) 设直线 $l$ 与 $x$ 轴的交点为 $N$, 则 $N$ 的坐标为 $(-1,0)$,

$\therefore S_{\triangle A O B}=\frac{1}{2}|O N| \cdot\left|y_{1}-y_{2}\right|$

$=\frac{1}{2} \times|O N| \times \sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} \cdot y_{2}}$

$=\frac{1}{2} \times 1 \times \sqrt{\frac{1}{k^{2}}+4}=\sqrt{10}$,
解得 $k^{2}=\frac{1}{36}$, 所以 $k= \pm \frac{1}{6}$.